可汗学院数学题目-24Ratios,ratesandproportions

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Morita classes of algebras in modular tensor categories

a rXiv:078.1897v2[mat h.CT]29J ul28KCL-MTH-07-100708.1897[math.CT]Morita classes of algebras in modular tensor categories Liang Kong a ∗and Ingo Runkel b †a Max-Planck-Institut f¨u r Mathematik Vivatsgasse 7,53111Bonn,Germany b Department of Mathematics,King’s College London Strand,London WC2R 2LS,United Kingdom August 2007Abstract We consider algebras in a modular tensor category C .If the trace pairing of an algebra A in C is non-degenerate we associate to A a commutative algebra Z (A ),called the full centre,in a doubled version of the category C .We prove that two simple algebras with non-degenerate trace pairing are Morita-equivalent if and only if their full centres are isomorphic as algebras.This result has an interesting interpretation in two-dimensional rational conformal ﬁeld theory;it implies that there cannot be several incompatible sets of boundary conditions for a given bulk theory.1Introduction and summaryIt is well-known that two Morita equivalent rings have isomorphic centres(see e.g.[AF,§21]). The converse is in general not true,a counter example is provided by the real numbers andthe quaternions.On the other hand,for simple algebras over C(or any algebraically closedﬁeld)the converse holds trivially,since all such algebras are of the form Mat n(C)and all have centre C.The situation becomes much richer if instead of considering algebras only in the category ofﬁnite-dimensional C-vector spaces one allows for more general tensor categories.Forexample,for the categories of integrable highest weight representations of the aﬃne Liealgebras sl(2)k,k=1,2,...,oneﬁnds an ADE-pattern for the Morita-classes,see e.g.[O]. These representation categories are in fact examples of so-called modular tensor categories,which are the class of categories we are considering in this paper.We call an algebra non-degenerate if the trace pairing(or rather the appropriate cat-egorical formulation thereof)is non-degenerate.We prove in this paper that two simple non-degenerate algebras in a modular tensor category are Morita equivalent if and only if they have isomorphic‘full centres’.The latter is a commutative algebra which is a general-isation of the centre of an algebra over C,but which typically lives in a diﬀerent category than the algebra itself.Our motivation to study the relation between Morita classes of algebras and their centrescomes from two-dimensional conformalﬁeld theory.It has recently become clear that there is a close relationship between rational CFT and non-degenerate algebras in modular tensor categories,both in the Euclidean and the Minkowski formulation of CFT,see e.g.[KLM, Fu1,LR,H2,Kg3,Fj3].In the Euclidean setting,the modular tensor category arises as the category of representations of a vertex operator algebra with certain additional properties [H1,H2],which we will refer to as‘rational’.The non-degenerate algebra A then is an algebra of boundaryﬁelds[Fu1],i.e.an open-string vertex operator algebra[HK1].It turns out that A and the rational vertex operator algebra together uniquely determine a CFT [Fu1,Fj1,Fj3];however,to ensure its existence,some complex analytic and convergence issues remain to be settled.As a consequence of the uniqueness,one can obtain from A the algebra of bulkﬁelds[Fj3],i.e.a fullﬁeld algebra[HK2].An important question then is if two non-Morita equivalent open-string vertex operator algebras can give rise to the same fullﬁeld algebra,or–in more physical terms–if there may exist several incompatible sets of boundary conditions for a given bulk CFT.Our result implies that for a CFT which is rational(in the sense that its underlying vertex operator algebra is rational),this cannot happen.Recall that an algebra in a tensor category C with associatorαU,V,W and unit constraintsl U,r U is a triple A=(A,m,η)where A is an object of C,m(the multiplication)is a morphism A⊗A→A such that m◦(m⊗id A)◦αA,A,A=m◦(id A⊗m),andη(the unit)is a morphism 1→A such that m◦(id A⊗η)=id A◦r A and m◦(η⊗id A)=id A◦l A.We will only consider unital algebras.In the following we will also assume that all tensor categories are strict to avoid spelling out associators and unit constraints.In the same way one deﬁnes left–,right–,and bimodules.For example,given two algebrasA andB ,an A -B -bimodule is a triple X =(X,ρl ,ρr )where ρl :A ⊗X →X and ρr :X ⊗B →X are the representation morphisms;ρl obeys ρl ◦(m A ⊗id X )=ρl ◦(id A ⊗ρl )and ρl ◦(ηA ⊗id X )=id X ,and similar for ρr .Furthermore the left and right action commute,i.e.ρr ◦(ρl ⊗id B )=ρl ◦(id A ⊗ρr ).With the help of bimodules we can now deﬁne when an algebra is simple,namely when it is simple as a bimodule over itself,and when two algebras A ,B are Morita equivalent,namely when there exist an A -B -bimodule X and a B -A -bimodule Y such that X ⊗B Y ∼=A and Y ⊗A X ∼=B as bimodules.Let now C be a modular tensor category (see[T]and e.g.[BK]),i.e.a semisimple C -linear abelian ribbon category with End(1)=C id 1,having a ﬁnite number of isomorphism classes of simple objects and whose braiding obeys a certain nondegeneracy condition.(This deﬁnition is slightly more restrictive than the original one in [T].)We will express morphisms in ribbon categories with the help of the usual graphical notation [JS];our conventions are summarised in [Fj1,app.A.1].Given an algebra A we can deﬁne the morphism ΦA :A →A ∨as ΦA =A∨.(1.1)As in [Fu3]we call an algebra A non-degenerate iﬀΦA is invertible (the deﬁnition still makes sense in a tensor category with dualities).This generalises the condition that the trace pairing a,b →tr(a ·b )of a ﬁnite-dimensional algebra over a ﬁeld is non-degenerate.We will list some properties of non-degenerate algebras in section 2.1below.Given an algebra A in C ,the non-trivial braiding leads to two notions of centre,namely the left centre C l (A )and the right centre C r (A )of A [VZ,O,Fr1].Denoting the braiding of C by c U,V :U ⊗V →V ⊗U ,the left centre is the largest subobject C l (A )ιl →A such that m ◦c A,A ◦(ιl ⊗id A )=m ◦(ιl ⊗id A )and the right centre the largest subobject C r (A )ιr →A such that m ◦c A,A ◦(id A ⊗ιr )=m ◦(id A ⊗ιr ).We will give a formulation of the left centre of a non-degenerate algebra as the image of an idempotent in section 2.3below.The ﬁnal ingredient we need to state our main result is a doubled version of C ,namely C ⊠˜C .Here the category ˜C is obtained from C by replacing braiding and twist with their inverses,and the product C ⊠˜Cis the completion with respect to direct sums of C ×˜C (where the objects are pairs of objects in C and the Hom-spaces are tensor products of the two corresponding Hom-spaces in C ).C ⊠˜Cis again a modular tensor category.In fact,there is a notion of a ‘centre’Z of a tensor category,and for a modular tensor category C one ﬁnds Z (C )∼=C ⊠˜C [M¨u 1].Apart from the tensor unit,the category C ⊠˜Ccontains another canonically given com-mutative non-degenerate algebra,deﬁned on the object R =⊕i ∈I U i ×U ∨i [M¨u 1,Fr1,Kg1].Here the (ﬁnite)set I indexes a choice of representatives U i of the isomorphism classes of simple objects in C .The multiplication and further properties of R are given in section 2.2.For a non-degenerate algebra A in C we can now deﬁne the full centre Z(A)as the leftcentre of the algebra(A×1)⊗R in C⊠˜C[Fj2,Fj3];our convention for the tensor product of algebras and some properties of the full centre will be discussed in section2.3.As opposed to the left and right centres,the full centre is not a subobject of A,in fact it is not evenan object of the same category.On the other hand,one can recover C l(A)and C r(A)from Z(A)by applying suitable projections.Furthermore,if C is the category V ect f(C)ofﬁnite-dimensional complex vector spaces then also C⊠˜C∼=V ect f(C),and the notions of left,right and full centre coincide and agree with the usual deﬁnition of the centre of an algebra over aﬁeld.The full centre turns out to be a Morita-invariant notion and our main result is that it can be used to distinguish Morita-classes of non-degenerate algebras.Theorem1.1Let C be a modular tensor category and let A,B be simple non-degenerate algebras in C.Then the following two statements are equivalent.(i)A and B are Morita equivalent.(ii)Z(A)and Z(B)are isomorphic as algebras.Remark1.2(i)In the special case C=V ect f(C)a simple non-degenerate algebra is isomorphic to thefull matrix algebra Mat n(C)for some n,and the full centre Z is just the usual centre,which in the case of Mat n(C)is C.The above theorem then just states that any two full matrix algebras over C are Morita equivalent.(ii)An algebra is called haploid iﬀdim Hom(1,A)=1[FS].Denote by C max(C⊠˜C)theset of isomorphism classes[B]of haploid commutative non-degenerate algebras B in C⊠˜C which obey in addition dim(B)=Dim(C),where Dim(C)= i∈I dim(U i)2.(It follows from[KO,thm.4.5]that this is the maximal dimension such an algebra can have.)Notethat[R]∈C max(C⊠˜C),with R deﬁned as above.Let further M simp(C)be the set of Morita classes{A}of simple non-degenerate algebras A in C.We will see in remark3.4(ii)that the assignment z:{A}→[Z(A)]is a well-deﬁned map from M simp(C)to C max(C⊠˜C).For example,z({1})=[R].Theorem1.1shows that z is injective.A result recently announced by M¨u ger[M¨u2]shows that z is also surjective.(An independent proof of surjectivity has subsequently appeared in[KR,sect.3.3].)(iii)A closed two-dimensional topologicalﬁeld theory is the same as a commutative Frobeniusalgebra B over C,see e.g.[Kk].In the case that B is semi-simple,the possible boundary conditions for the theory deﬁned by B can be classiﬁed by K0(B-mod)[Mo,MS].For a (rational)two-dimensional conformalﬁeld theory the boundary conditions can be classiﬁed by K0(A-mod)where A is a non-degenerate algebra in C,and C in turn is the representation category of a rational vertex algebra V[Fu1].The algebra A comes from the boundaryﬁelds –i.e.from an open-string vertex algebra over V–for one of the possible boundary conditions [Fu1,HK1,Kg2].For the topological theory,the category C is given by C=V ect f(C)and for B one can choose the centre of A.(If A is not simple this choice is not unique,see[LP] and[Fj3,rem.4.27].)For V ect f(C),A and B=Z(A)are Morita-equivalent,and so K0of A-mod and B-mod agree.In general oneﬁnds that,for A a simple non-degenerate algebrain a modular tensor category C and B=Z(A)the full centre,# isocl.of simple B-left modules in C⊠˜C =# isocl.of simple A-A-bimodules C . This can be computed from[Fu1,thm.5.18]together with the fact that Z(A)has a unique(up to isomorphism)simple local left module,namely Z(A)itself.Thus in general,K0(B-mod)–the Grothendieck group of the category of B-left modules in C⊠˜C–is related to defect lines (see[Fu1,rem.5.19]and[Fr2]),and its relevance for the classiﬁcation of boundary conditions is special to the topological case.Nonetheless,there is a connection between B and boundary conditions:We will see in section4that via the tensor functor T:C⊠˜C→C one obtains an algebra T(B)in C which is a direct sum of simple non-degenerate algebras,all of which are Morita-equivalent to A.In fact(cf.prop.4.3below)one has that K0(T(B)-mod)∼= K0 A-mod ×n,where n is the number of isomorphism classes of simple A-left modules in C.The rest of the paper is organised as follows.In section2we collect some results on non-degenerate algebras and the full centre.Section3we prove that statement(i)in theorem 1.1implies(ii)and in section4we prove the converse.2Preliminaries2.1Properties of non-degenerate algebrasNot all the properties discussed in this section require us to work with the full structure of a modular tensor category and we therefore state them in the appropriate context.However, all these properties do in particular hold for modular tensor categories.Let C be a(strict)tensor category.In the same way that one deﬁnes an algebra in C one can deﬁne a coalgebra A=(A,∆,ε)where∆:A→A⊗A andε:A→1obey co-associativity and the counit condition.Deﬁnition2.1A Frobenius algebra A=(A,m,η,∆,ε)is an algebra and a coalgebra such that the coproduct is an intertwiner of A-bimodules,i.e.(id A⊗m)◦(∆⊗id A)=∆⊗m= (m⊗id A)◦(id A⊗∆).We will use the following graphical representation for the morphisms of a Frobenius algebra,m=,η=,∆=,ε=.(2.1)A Frobenius algebra A in a k-linear tensor category,for aﬁeld k,is called special iﬀm◦∆=ζid A andε◦η=ξid1for nonzero constantsζ,ξ∈k.Ifζ=1we call A normalised-special.A(strictly)sovereign tensor category is a tensor category equipped with a left and a right duality which agrees on objects and morphisms(see e.g.[B,FS]for more details).Wewill write the dualitiesas=d U:U∨⊗U→1,=˜d U:U⊗U∨→1,∨=b U:1→U⊗U∨,∨=˜b U:1→U∨⊗U,(2.2)(n.b.,‘b’stands for birth and‘d’for death).Given these dualities one can deﬁne the left and right traces of a morphism f:U→U as tr l(f)=d U◦(id U∨⊗f)◦˜b U and tr r(f)=˜dU◦(f⊗id U∨)◦b U,as well as theleft and right dimension of U,dim l/r(U)=tr l/r(id U).If U∼=U∨,then dim l(U)=dim r(U)[FS,rem.3.6.3].In a modular tensor category(and more generally in a spherical category)the left and right traces and dimensions always coincide.Let now C be a sovereign tensor category.A Frobenius algebra in C is symmetric iﬀ∨=∨.(2.3)For a Frobenius algebra(symmetric or not)one has dim l(A)=dim r(A)since the two morphisms in(2.3)are in fact isomorphisms(see e.g.the proof of lemma3.7in[Fu1]),and so A∼=A∨.We will write dim(A)≡dim l/r(A).In a k-linear sovereign category with End(1)=k id1we will identify dim l/r(U)with the corresponding element of k via tr l/r(id U)=dim l/r(U)id1.In this case oneﬁnds that for a normalised-special symmetric Frobenius algebra one hasε◦η=dim(A)id1[FS,sect.3];in particular,dim(A)=0.Deﬁnition2.2An algebra A in C is non-degenerate if the morphismΦA in(1.1)is invertible.The relation between non-degenerate algebras and Frobenius algebras is summarised in the following lemma.Lemma2.3Let C be a sovereign tensor category.(i)Let A be a non-degenerate algebra in C.Taking∆=(Φ−1A⊗m)◦(˜b A⊗id A)andε=η∨◦ΦA turns A into a symmetric Frobenius algebra which obeys m◦∆=id A andε◦η= dim(A).(ii)Let A be a symmetric Frobenius algebra in C such that m◦∆=id A.Then A is a non-degenerate algebra.(iii)Two non-degenerate algebras A and B are isomorphic as algebras if and only if they are isomorphic as Frobenius algebras(with counit and coproduct as given in(i)).(iv)If C is in addition k-linear with End(1)=k id1and A is a non-degenerate algebra in C with dim(A)=0,then A is special.This lemma can be proved by combining and adapting lemmas3.7,3.11and3.12of[Fu1].Part of the proof of(i)involves showing thatΦA in(1.1)is also equal to the morphism obtained by‘reﬂecting’the graph along a vertical axis(cf.[Fu1,eqn.(3.33)]),and thatequally∆=(m⊗Φ−1A )◦(b A⊗id A).In this sense,the Frobenius algebra structure on anon-degenerate algebra does not involve any arbitrary choices.Whenever we will consider a non-degenerate algebra as a Frobenius algebra we mean the coproduct and counit given in part(i)of the above lemma.In the setting we will work with below,C is a modular tensor category and one can convince oneself that a simple non-degenerate algebra in C necessarily has dim(A)=0,cf. [Fr2,lem.2.6].In particular,a simple non-degenerate algebra is then always also normalised-special symmetric Frobenius.Let now C be an abelian sovereign tensor category.Let A be a non-degenerate algebra in C and let M be a right A-module and N be a left A-module.The tensor product M⊗A N can be written as the image of the idempotentP⊗A=ρM N(2.4)That is,there exist morphisms e A:M⊗A N→M⊗N and r A:M⊗N→M⊗A N such thatr A◦e A=id M⊗A N and e A◦r A=P⊗A.One can convince oneself that r A:M⊗N→M⊗A Nfulﬁls the universal property of the coequaliser ofρM⊗id N and id M⊗ρN.2.2Modular tensor categoriesLet C be a modular tensor category.Recall from section1that we chose representatives {U i|i∈I}for the isomorphism classes of simple objects.We alsoﬁx U0=1and for an index k∈I we deﬁne the index¯k by U¯k∼=U∨k.The numbers s i,j∈C are deﬁned vias i,j id1=tr(c Ui,U j ◦c Uj,U i).(2.5)They obey s i,j=s j,i and s0,i=dim(U i),see e.g.[BK,sect.3.1].The non-degeneracy condition on the braiding of a modular tensor category is that the|I|×|I|-matrix s should be invertible.In fact,k∈Is ik s kj=Dim(C)δi,¯ (2.6)(cf.[BK,thm.3.1.7]),where as above Dim(C)= i∈I dim(U i)2.In particular,Dim(C)=0. One can show(even in the weaker context of fusion categories over C)that Dim(C)≥1 [ENO,thm.2.3].Let usﬁx a basis{λα(i,j)k }N k ijα=1in Hom(U i⊗U j,U k)and the dual basis{Υ(i,j)kα}N k ijα=1inHom(U k,U i⊗U j).The duality of the bases means thatλα(i,j)k ◦Υ(i,j)kβ=δα,βid Uk.We alsoﬁx λ(0,i )i =λ(i,0)i =id U i .We denote the basis vectors graphically as follows:λα(i,j )k=i j ,Υ(i,j )k α=k j.(2.7)As in section1let R be the object in C ⊠˜C given by R =⊕i ∈I U i ×U ∨i .We deﬁne a unit morphism ηR :1×1→R to be the natural embedding and a multiplication morphismm R :R ⊗R →R asm R = i,j,k ∈I N kij α=1×i j∨i j .(2.8)The basis elements λα(i,j )k and their duals enter the above expression in such a way that m R is independent of the choice of ing (2.8),one can compute ΦR (deﬁned in (1.1))explicitly,resulting in ΦR =Dim(C ) i ∈I f i ×g i where f i :U i →U ∨¯ıand g i :U ∨i →U ∨∨¯ıare given by f i =(λ1(i,¯ı)0⊗id U ∨¯ı)◦(id U i ⊗b U ¯ı),g i =(δU ¯ı⊗˜d U i )◦ (c −1U ¯ı,U i◦Υ(i,¯ı)01)⊗id U ∨i ,(2.9)and δU :U →U ∨∨is the isomorphism (˜dU ⊗id U ∨∨)◦(id U ⊗b U ∨).It follows from [M¨u 1,prop.4.1](see also [Fr1,lem.6.19]and [Kg1,thm.5.2])that the three morphisms ηR ,m R and ΦR give R the structure of haploid commutative non-degenerate algebra.Thus it is also normalised-special symmetric Frobenius.(The algebra R can also be deﬁned in more general categories,see [M¨u 1,Fr1].)2.3Properties of the full centreFrom hereon we will always take C to be a modular tensor category.Most of the constructions in this section can be carried out in greater generality,see e.g.[Fr1],but for the purpose of the proof of theorem 1.1this will not be necessary.An algebra A in a braided tensor category has a left centre and a right centre [VZ,O],both of which are sub-algebras of A .We will only need the left centre.The following deﬁnition is the one used in [Fr1],which in our setting is equivalent to that of [VZ,O].Deﬁnition 2.4The left centre C l (A )of a non-degenerate algebra A in C is the image of the idempotent P l (A ):A →A ,whereP l (A )=.(2.10)That is,there are morphisms C l(A)ιl→A and A r l→C l(A)such that r l◦ιl=id Cl(A)andιl◦r l=P l(A).This follows from deﬁnition2.31and remark2.34of[Fr1].(In[Fr1]the algebra A is assumed to be special,but the relevant proofs only use m◦∆=id A,which is satisﬁed by a non-degenerate algebra according to lemma2.3(i).)The proof of the following lemma can be found in[Fr1,prop.2.37].Lemma2.5Let A be a non-degenerate algebra in a modular tensor category C.(i)C l(A)is a commutative symmetric Frobenius algebra.(ii)If C l(A)is simple and dim(C l(A))=0,then C l(A)is in addition special.Given two algebras A and B we deﬁne a multiplication on the tensor product A⊗Bas m A⊗B=(m A⊗m B)◦(id A⊗c−1A,B ⊗id B)and a unit morphism asηA⊗B=ηA⊗ηB.Thisturns A⊗B into an algebra.Note that one can also deﬁne a diﬀerent multiplication m′A⊗Bby using c B,A instead of c−1A,B .The resulting algebra is isomorphic to(A op⊗B op)op,where‘op’stands for the opposed algebra,see[Fu1,rem.3.23].We will always use m A⊗B.For two coalgebras we similarly set∆A⊗B=(id A⊗c A,B⊗id B)◦(∆A⊗∆B)andεA⊗B=εA⊗εB.This turns A⊗B into a coalgebra.One easily checks that if A and B share any of the properties non-degenerate,Frobenius,symmetric,special,then the property is inherited by A⊗B.On the other hand,even if A and B are commutative,A⊗B is generally not.For an object U of C denote by R(U)the object in C⊠˜C given by R(U)=(U×1)⊗R. (R(·)can be understood as the adjoint of the functor T mentioned in remark1.2(iii);more details can be found in[KR,sect.2.4].)If A is a non-degenerate algebra in C then A×1is a non-degenerate algebra in C⊠˜C and the above discussion gives a non-degenerate algebra structure on R(A).Deﬁnition2.6[Fj3,def.4.9]The full centre Z(A)of A is deﬁned to be C l(R(A)).Proposition2.7Let A be a non-degenerate algebra in a modular tensor category C.(i)Z(A)is a commutative symmetric Frobenius algebra with dim(Z(A))=d·Dim(C)for some integer d≥1.(ii)If A is simple then Z(A)is a haploid commutative non-degenerate algebra with dim(Z(A)) =Dim(C).Furthermore,Z(A)is normalised-special.Proof.Theﬁrst statement in part(i)follows from lemma2.5(i)together with the above observation that R(A)is a non-degenerate algebra in C⊠˜C.For the statement about the dimension,let Z ij=dim Hom(Z(A),U i×U j).By combining[Fj2,eqn.(A.3)](note that in[Fj2]Z(A)has a diﬀerent meaning,namely the object given in eqn.(3.9)there)with eqn.(5.65)and theorem5.1of[Fu1]it follows that k∈I Z ik s kj= l∈I s il Z ing this we can computedim(Z(A))= i,j Z ij dim(U i)dim(U j)= i,j s0i Z ij s j0= j,k Z0k s kj s j0= k Z0kδk,0Dim(C)=Z00Dim(C).(2.11)It follows from the equalities(A.2)in[Fj2]that Z00=dim Hom A|A(A,A),where Hom A|A(·,·) denotes the space of bimodule intertwiners.Since id A is a bimodule intertwiner we have Z00≥1.For(ii)note that in the present setting,A is simple iﬀit is absolutely simple,i.e.iﬀHom A|A(A,A)=C id A,which is equivalent to Z00=1.Therefore,A is simple iﬀZ(A) is haploid.Since by assumption in(ii),A is simple,(2.11)holds with Z00=1.Recall from above that Dim(C)≥1,so that altogether we see that Z(A)is simple(since it is haploid) and has nonzero dimension.By lemma2.5(ii),Z(A)is then also special.We can rescale the coproduct(and the counit)to make Z(A)normalised-special and it then follows from lemma2.3(ii)that Z(A)is non-degenerate.Using Q X we deﬁne a morphism D X:Z(B)=C l((B×1)⊗R)→Z(A)=C l((A×1)⊗R) by composing with the corresponding embedding and restriction morphisms,D X=r l◦Q X◦ιl.(3.2) As a direct consequence of lemma3.1we have D X=D X′for two isomorphic bimodules X and X′,as well as,for X,Y as in lemma3.1,D A=id Z(A),D X◦D Y=D X⊗B Y.(3.3) Lemma3.2Let A,B be non-degenerate algebras(not necessarily simple)and X an A-B-bimodule,such that dim(A),dim(B)and dim(X)are non-zero and theidentitiesA=dim(X)dim(B)id B,(3.4)and∨=dim(A)dim(B)D X:Z(B)→Z(A)is an isomorphism of Frobenius algebras.The precise form of the dimension-factors appearing in conditions(3.4)and(3.5)is not an extra condition,but is in fact uniquelyﬁxed.For example composing theﬁrst equation in(3.4)withεA from the left andηA from the right gives theﬁrst constant.Also note that X∨is naturally a B-A-bimodule,see e.g.[Fr2,sect.2.1].Proof of lemma3.2a)A∼=X⊗B X∨as A-A-bimodules:We deﬁne two morphisms f1:A→X⊗B X∨and f2:X⊗B X∨→A byf1=dim(A)c)φX is an isomorphism :First note that taking the trace of(3.5)and using (3.4)results in the identitydim(X )2=dim(A )dim(B ).(3.7)Using this,as well as (3.3)and part b)we obtainφX ∨◦φX =dim(X )dim(B )D X ∨◦D X =D X ∨⊗A X =D B =id Z (B ),(3.8)In the same way one checks that φX ◦φX ∨=id Z (A ).Thus φX is an isomorphism.d)φX is an algebra map :The unit property φX ◦ηZ (B )=ηZ (A )can be seen as follows,dim(X )dim(B )A ×Z (A )=Z (A )r l(3.9)The compatibility with the multiplication,m Z (A )◦(φX ⊗φX )=φX ◦m Z (B ),amounts to the identitiesdim(X )dim(B )2Z (A )(2)=dim(X )3dim(B)Z (A )ιl (3.10)The left hand side is obtained by writing out the deﬁnitions of the various morphisms in m Z (A )◦(φX ⊗φX).In step (1)the two projectors ιl ◦r l =P l (R (A ))have been omitted using lemma 3.1(ii,iv),and the uppermost multiplication morphism of A has been replaced by a representation morphism of the bimodule X .In step (2)we used property (3.5).For step (3)note that the B ×1-ribbon connecting X ×1to itself can be rearranged (using that B is symmetric Frobenius,as well as the representation property)to the projector P l (R (B ))which can be omitted against ιl .Using the representation property on the re-maining two B ×1-ribbons,as well as (3.7),gives the right hand side of (3.10).Replac-ing Q X =Q X ◦P l (R (B ))=Q X ◦ιl ◦r l ﬁnally shows that the right hand side is equal to φX ◦m Z (B ).e)φX is a coalgebra map :For this part of the statement,the coproduct and counit of Z (A )and Z (B )have to be normalised as in the proof of [Fr1,prop.2.37].That is,while the mul-tiplication and unit on Z (A )are given by m Z (A )=r l ◦m R (A )◦(ιl ⊗ιl )and ηZ (A )=r l ◦ηR (A ),for the coproduct and counit we choose∆Z (A )=ζ−1(r l ⊗r l )◦∆R (A )◦ιl ,εZ (A )=ζεR (A )◦ιl ,ζ=dim(Z (A ))dim(X )∨,(3.12)where in the second step (3.5)is substituted.One also needs to use that dim(Z (A ))=dim(Z (B )),which follows from part c).This completes the proof of the lemma.Proof of(i)⇒(ii)in theorem1.1:By assumption the simple non-degenerate algebras A and B are Morita-equivalent.Therefore there exists an A-B-bimodule X and a B-A-bimodule Y such that A∼=X⊗B Y and B∼= Y⊗A X as bimodules.Lemma3.3ensures that the conditions of lemma3.2are met.Thus the morphismφX:Z(B)→Z(A)is an isomorphism of algebras.l−11.For U,V,W,X∈C,notice thatT U×V ⊗T W×X =(U⊗V)⊗(W⊗X),T (U×V)⊗(W×X) =(U⊗W)⊗(V⊗X).(4.1) We deﬁneϕ2:T U×V ⊗T W×X →T (U×V)⊗(W×X) byϕ2:=αU,W,V⊗X◦ id U⊗α−1W,V,X ◦ id U⊗(c−1W V⊗id X) ◦(id U⊗αV,W,X)◦α−1U,V,W⊗X.(4.2) The above deﬁnition ofϕ2can be naturally extended to a morphism T(M1)⊗T(M2)→T(M1⊗M2)for any pair of objects M1,M2in C⊠˜C.We still denote the extended morphism asϕ2.(We hide the dependence ofϕ2on M1,M2in our notation for simplicity.)We have Lemma4.1The functor T together withϕ0andϕ2is a tensor functor.Note that T takes algebras to algebras(see for example[Kg2,prop.3.7])but in general does not preserve commutativity.Explicitly,if(B,m B,ηB)is an algebra in C⊠˜C,then the triple(T(B),m T(B),ηT(B)),wherem T(B):=T(m B)◦ϕ2,ηT(B):=T(ηB)◦ϕ0,(4.3) is an algebra in C.4.2The full centre transported to C and simple modulesLet now C again be a(strict)modular tensor category,and let A be a simple non-degenerate algebra in C.As observed in section1,this implies in particular that dim(A)=0.The category of left A-modules is again semisimple and abelian[FS,props.5.1and5.24]with a ﬁnite number of isomorphism classes of simple objects(this follows e.g.by combining the fact that C itself only has aﬁnite number of isomorphism classes of simple objects with[FS, lem.4.15]).Let{Mκ|κ∈J}be a set of representatives of the isomorphism classes of simple left A-modules.Lemma4.2Let A be a non-degenerate algebra in C and let M be a left A-module.(i)M∨⊗A M is an algebra with unit e A◦˜b M and multiplication r A◦(id M∨⊗˜d M⊗id M)◦(e A⊗e A).(ii)M is simple if and only if M∨⊗A M is haploid.Proof.Part(i)is a straightforward calculation,see e.g.[Fu2,eqn.(2.48)].Claim(ii)follows since Hom A(M,M)∼=Hom(M∨⊗A M,1).Theﬁrst space is one-dimensional iﬀM is simple, and the second space is one-dimensional iﬀM∨⊗A M is haploid.。

2018年ACT数学练习题及答案解析

2018年ACT数学练习题及答案解析本文主要为考生整理了ACT数学5道选择题，并对这5道选择题进行了详细解析。

除练习题外，我们还为考生整理了9点ACT 数学答题技巧，详情如下：1、3x3·2x2y·4x2y is equivalent toA. 9x7y2B. 9x12y2C. 24x7y2D. 24x12yE. 24x12y2正确答案选C。

常数相乘(3·2·4)，合并x项(x3x2x2 → x3+2+2 → x7) 和y项(y·y → y1y1 → y1+1→ y2)，结果为24x7y2。

解析：The correct answer is C. To find an equivalent expression, you can multiply the constants (3·2·4), combine the x term (x3x2x2 → x3+2+2 → x7, because when you have a common base you use the base and add the exponents), and combine the y t erms (y·y → y1y1 → y1+1 → y2). The result is 24x7y2.2、 Mr. Dietz is a teacher whose salary is $22,570 for this school year, which has 185 days. In Mr. Dietz's school district, substitute teachers are paid $80 per day. If Mr. Dietz's takes a day off without pay and a substitute teacher is paid to teach Mr. Dietz's classes, how much less does the school district pay in salary by paying a substitute teacher instead of paying Mr. Dietz's for that day?F. $ 42G. $ 80H. $ 97J. $105K. $122正确答案选F。

新gre数学知识点

新gre数学知识点新GRE数学知识点一、算术（Arithmetic）1. 整数（Integers）- 奇数与偶数- 质数与合数- 整数的四则运算2. 分数、小数和百分比（Fractions, Decimals, and Percentages） - 分数的加减乘除- 小数与分数的互转- 百分比的计算与转换3. 比率和比例（Ratios and Proportions）- 比率的概念与计算- 比例的概念与计算- 交叉相乘法二、代数（Algebra）1. 线性方程和不等式（Linear Equations and Inequalities）- 一元一次方程- 一元一次不等式- 系统线性方程组2. 二次方程和不等式（Quadratic Equations and Inequalities） - 二次方程的标准形式- 配方法- 完全平方- 二次不等式的解法3. 函数及其性质（Functions and their Properties） - 函数的定义与表示- 线性函数、二次函数、绝对值函数- 函数的变换：平移、伸缩、对称三、几何（Geometry）1. 平面几何（Plane Geometry）- 点、线、面的基本性质- 三角形、四边形的性质与计算- 圆的性质与计算2. 空间几何（Solid Geometry）- 多面体的性质与计算- 圆柱、圆锥、球的性质与计算3. 坐标几何（Coordinate Geometry）- 坐标系中点的表示- 距离公式、斜率公式- 线性方程的图形表示四、数据分析（Data Analysis）1. 数据的描述（Descriptive Statistics）- 平均数、中位数、众数- 方差、标准差- 百分位数2. 概率（Probability）- 事件的概率- 条件概率与独立事件- 排列组合的基本公式3. 样本与假设检验（Samples and Hypothesis Testing）- 样本的基本概念- 假设检验的基本步骤- 置信区间的概念以上是新GRE数学部分的主要知识点概览。

可汗学院数学题目-41Complexnumbers[1]

Complex numbers1.(−8+4i)(1−i)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.−12+4iB.−12+12iC.−4+12iD.−4+4iCorrect answer: C Difficulty level: 22.(4+i)2Which of the following is equivalent to **plex number shown above?Note: i=√−1A.15+8iB.15−8iC.17+8iD.17−8iCorrect answer: A Difficulty level: 23.(8−2i)(4−2i)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.28−24iB.28+8iC.36−24iD.36+8iCorrect answer: A Difficulty level: 24.(5+i)(7−3i)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.32+8iB.32−8iC.38+8iD.38−8iCorrect answer: D Difficulty level: 25.i4+4i2+4Which of the following is equivalent to **plex number shown above?Note: i=√−1A.1B.−1C.i+4D.i−4Correct answer: A Difficulty level: 26.(−3−i)(4−2i)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.−14−2iB.−14+2iC.−10−2iD.−10+2iCorrect answer: B Difficulty level: 27.(6+2i)2Which of the following is equivalent to **plex number shown above?Note: i=√−1A.40+4i2B.40+24iC.32+24iD.32+4i2Correct answer: C Difficulty level: 28.(1+i)(1−i)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.2-2iB.2iC.0D.2Correct answer: D Difficulty level: 29.i(7−3i)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.4iB.10iC.7i−3D.7i+3Correct answer: D Difficulty level: 210.i2−16i+4Which of the following is equivalent to **plex number shown above?Note: i=√−1A.i−4B.i+4C.−i−4D.−i+4Correct answer: A Difficulty level: 211.(3+i)(2−4i)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.2−10iB.2−14iC.10−10iD.10−14iCorrect answer: C Difficulty level: 212.i101Which of the following is equivalent to **plex number shown above?Note: i=√−1A.1B.−1C.iD.−iCorrect answer: C Difficulty level: 213.(5−i)2Which of the following is equivalent to **plex number shown above?Note: i=√−1A.24−10iB.24+10iC.26−10iD.26+10iCorrect answer: A Difficulty level: 214.−8(7i−3i2)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.−80iB.−56i−24C.−56+24iD.−32iCorrect answer: B Difficulty level: 315.32+iWhich of the following is equivalent to **plex number shown above?Note: i=√−1A.2−iB.2+iC.6+3i5D.6+3i5Correct answer: D Difficulty level: 316.(3−i)3Which of the following is equivalent to **plex number shown above?Note: i=√−1A.8−26iB.18−26iC.27−26iD.30−26iCorrect answer: B Difficulty level: 317.(5−7i+i2)+(8i3+12)**plex expression above is equivalent to the expression a+bi for the integer constants a and b.What is the value of a?Note: i=√−1A.16B.17C.18D.19Correct answer: A Difficulty level: 318.(−3+2i)(1−i3)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.−5−iB.−5+5iC.−1−iD.−1+5iCorrect answer: A Difficulty level: 319.i11+i13Which of the following is equivalent to **plex number shown above?Note: i=√−1A.−2iB.2iC.0D.2Correct answer: C Difficulty level: 320.51+3iWhich of the following is equivalent to **plex number shown above?Note: i=√−1A.1+3i2B.1−3i2C.−5(1+3i)8D.−5(1−3i)8Correct answer: B Difficulty level: 321.(10−8i3)−(6+i)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.4−7iB.4+7iC.4+9iD.4−9iCorrect answer: B Difficulty level: 322.11−iWhich of the following is equivalent to **plex number shown above?Note: i=√−1A.2−2iB.2+2iC.1−i2D.1+i2Correct answer: D Difficulty level: 323.21−iWhich of the following is equivalent to **plex number shown above?Note: i=√−1A.1−iB.1+iC.2−iD.2+iCorrect answer: B Difficulty level: 324.8ix=−5What is the value of x in the equation above?Note:i=√−1A.−8i5B.8i5C.−5i8D.5i8Correct answer: D Difficulty level: 325.(2−3i)3Which of the following is equivalent to **plex number shown above?Note: i=√−1A.−46−9iB.−26−9iC.26−9iD.46−9iCorrect answer: A Difficulty level: 326.(12+i)(8−6i)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.−2+5iB.2+2iC.10+5iD.14+2iCorrect answer: C Difficulty level: 327.11−6i −11+6iWhich of the following is equivalent to **plex number shown above? Note: i=√−1A.1237iB.−1237iC.1237D.−1237Correct answer: A Difficulty level: 328.(23+12i)(12−13i)**plex expression above is equivalent to the expression a+bi for the rational constants a and b. What is the value of b?Note: i=√−1A.b=16B.b=−16C.b=496D. b=529Correct answer: D Difficulty level: 329.P(x)=2x2+3x−17If x=8−2i, what is the value of the polynomial P above?Note:i=√−1A.15−2iB.23−6iC.127−70iD.135−62iCorrect answer: C Difficulty level: 330.(6+i2)(2−2i)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.6−8iB.8−8iC.10−8iD.12−8iCorrect answer: B Difficulty level: 331.3i +2i2Which of the following is equivalent to **plex number shown above?Note: i=√−1A.3i+2B.3i−2C.−3i+2D.−3i−2Correct answer: D Difficulty level: 332.21+iWhich of the following is equivalent to **plex number shown above?Note: i=√−1A.−1+iB.−1−iC.1+iD.1−iCorrect answer: D Difficulty level: 433.3i10+i11Which of the following is equivalent to **plex number shown above?Note: i=√−1A.3+iB.−3+iC.3−iD.−3−iCorrect answer: D Difficulty level: 434.P(n)=n2−5n−7What is the value of P(−3i)?Note:i=√−1A.−4+15iB.−7+12iC.−7+24iD.−16+15iCorrect answer: D Difficulty level: 435.√3t2+5t+√27=0Which of the following is a solution to the equation above?Note:i=√−1A.t=−4√11i6B.t=−3√33i6C.t=−5+√11i6D.t=−5√3+√33i6Correct answer: D Difficulty level: 436.29=3(x+7)2+41Which of the following is a solution to the equation above?Note:i=√−1A.x=−7+2iB.x=−42−12iC.x=−76+√4436iD.x=−7−√1233iCorrect answer: A Difficulty level: 437.(8−2i)2(8+2i)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.60B.68C.480−120iD.544−136iCorrect answer: D Difficulty level: 438.√54i41√27i101Which of the following is equivalent to **plex number shown above?Note: i=√−1A.−√2iB.−√2C.√2D.√2iCorrect answer: C Difficulty level: 439.22−i −22+iWhich of the following is equivalent to **plex number shown above? Note: i=√−1A.4i5B.−4i5C.2i3D.−2i3Correct answer: A Difficulty level: 440.1+2i1−2i ÷1−2i1+2iWhich of the following is equivalent to **plex number shown above? Note: i=√−1A.1B.−1C.−725+2425iD.−725−2425iCorrect answer: D Difficulty level: 441.i3+i2Which of the following is equivalent to **plex number shown above?Note: i=√−1A.−1B.−2C.−1+iD.−1−iCorrect answer: D Difficulty level: 442.m2+6m+10=0Which of the following are solutions to the equation above?I. -3+iII. -3-iIII. 3+iNote:i=√−1A.I onlyB.I and II onlyC.I and III onlyD.I, II, and IIICorrect answer: B Difficulty level: 443.704i1776Which of the following is equivalent to **plex number shown above?Note: i=√−1A.704B.−704C.704iD.−704iCorrect answer: A Difficulty level: 444.12+5i −4+3i3−iWhich of the following is equivalent to **plex number shown above?Note: i=√−1A.10+26i(2+5i)(3−i)B.10−26i(2+5i)(3−i)C.10+27i(2+5i)(3−i)D.10−27i(2+5i)(3−i)Correct answer: D Difficulty level: 445.5+7i6−3iWhich of the following is equivalent to **plex number shown above?Note: i=√−1A.9+57i45B.9+57i3C.51+57i45D.51+57i3Correct answer: A Difficulty level: 446.9+7i184−i**plex expression above is equivalent to the expression a+bi for the rational constants a and b.What is the value of b?Note: i=√−1A.b=215B.b=217C.b=−1D.b=−7Correct answer: B Difficulty level: 447.5−i+(11−i)z=40+18iWhat is the value of z in the equation above?Note:i=√−1A.z=−19+24iB.z=24+20iC.z=3+2iD.z=3.2+145iCorrect answer: C Difficulty level: 448.2i+4ℎ−14=2iℎWhat is the value of h in the equation above?Note:i=√−1A.ℎ=3+iB.ℎ=72C.ℎ=7−iD.ℎ=83Correct answer: A Difficulty level: 4。

SAT考试备考数学题目24.Ratios, rates and proportions

考试复习LEVEL 31.Playing tennis Playing squash Energyexpenditurerate1,594௞௃௛ 2,726௞௃௛ Oxygenconsumptionrate77௅௛ 130௅௛ Fat burning rate 49 ௚௛ rThe table above lists average expenditure rates of playing tennis vs. playing squash based on a body weight of 55 kilograms. The ratio of each value in the tennis column to the corresponding value in the squash column is about the same throughout the table. Which of the following is closest to the fat burning rate r of squash?A. 27௚௛B. 29௚௛C. 83௚௛D. 85௚௛Correct answer: C2. Lea’s car travels an average of 30 miles per gallon of gas. If she spent $20.70 on gas for a 172.5 mile trip, what was the approximate cost of gas in dollars per gallon?A. $1.45 per gallonB.$3.40 per gallonC.$3.60 per gallonD.$5.75 per gallonCorrect answer: C3.Kavitha and Andrei are truck drivers for the Delightful Delivery Company,which services cities W, T, and Q.The distance from city W to city T is 48.5 miles.The distance from city T to city Q is 20 miles.The distance from city Q to city W is 36 miles.If Kavitha's truck and Andrei's truck travel at the same speed, and it takesAndrei 91 minutes to go from city W to city T, about how much time will it take Kavitha to travel from city W to city Q?A.19 minutesB.38 minutesC.68 minutesD.123 minutesCorrect answer: C4.Yasemin is using a sugar cookie recipe with a flour to sugar ratio of 11:6. Tomake one batch of cookies, 2.75 cups of flour are needed. If Yasemin would like to triple the recipe, which of the following best approximates how much sugar she will need?A.1.5 cupsB. 4.5 cupsC. 5 cupsD.15 cupsCorrect answer: B5. A certain high school geometry class is made up of freshmen and sophomores.The ratio of freshmen to sophomores in that class is 3:4. If thereare 12 sophomores in the class, what is the total number of students in the class?A.9B.16C.21D.28Correct answer: CLEVEL 41.The musical interval between two sounds is called an "octave" if the ratio of thesounds' frequencies is 2:1. The following table shows the names of the musical intervals between two sounds based on the ratios of the two sounds' frequencies. Name of the musical interval Ratio of two sound frequenciesMajor third 4:5Perfect fourth 3:4Perfect fifth 2:3Major sixth 3:5If a sound is played with a frequency of 480 Hz, and a second sound is played with a frequency of 800 Hz, what is the name of the musical interval between the two sounds?A.Major thirdB.Perfect fourthC.Perfect fifthD.Major sixthCorrect answer: D2.The "maximum occupancy" of a room is the total number of people who can be ina room without causing a fire hazard. In a large room, a Fire Safety Code statesthat the maximum occupancy is 1 person for every 7 square feet (fݐଶ). A college is hosting a concert in a hall that is 14,721 fݐଶ, and 1,000 people are expected to attend. According to the Fire Safety Code, approximately how many more people can attend the concert without causing a fire hazard?A.900 peopleB.1,000 peopleC.1,100 peopleD.2,100 people Correct answer: C3. A donut company makes cream-filled donuts using ଵସcup (c) of dough andଵଶtablespoon (tbsp) of cream per donut. The company decides to change their recipe to use 3 times the amount of cream for their "New Triple-Stuffed Donuts!" If the donut company's new recipe uses the same amount of dough per donut, what is the ratio of dough to cream needed to make 12 triple-stuffed donuts?A.1 c:3 tbspB. 1 c:6 tbspC. 2 c:3 tbspD.1 c:2 tbspCorrect answer: B4. A marine aquarium has a small tank and a large tank, each containing only red andblue fish. In each tank, the ratio of red fish to blue fish is 3 to 4. The ratio of fish in the large tank to fish in the small tank is 46 to 5. What is the ratio of blue fish in the small tank to red fish in the large tank?A.15:184B.10:69C.69:10D.184:15Correct answer: B5.Two leading brands of paper towels are on sale. Brand A has 6 rolls, eachwith 56 sheets, for $4.29. Brand B has 8 rolls, each with 48 sheets, for $5.99.Which of the following best describes the relationship between the cost per sheet of the two brands?A.The two brands cost the same amount per sheet.B.Brand B costs $0.003 more per sheet than Brand A.C.Brand A costs $0.003 more per sheet than Brand B.D.Brand B costs $0.03 more than Brand A.Correct answer: BLEVEL 21.Zhang Lei spent $20.00 during his last outing at the bowling alley. This included aone time shoe rental fee of $3.50. He spent the rest of the money on bowling a number of games. If it took Zhang Lei 45 minutes to bowl each game and hespent 2 hours and 15 minutes bowling, how much did it cost per game?A.$3.00B.$5.50C.$6.67D.$7.34Correct answer: B Level：22.Mr. Bumble delivers newspapers to his community every week. His routetakes 3 hours to complete. During the first 2 weeks of delivering newspapers, he traveled a total of 100 blocks. Mr. Bumble delivers newspapers for 50 weeks each year. How many blocks does Mr. Bumble travel delivering newspapers in 1 year?A.1 blockB. 4 blocksC.50 blocksD.2,500 blocksCorrect answer: D3.During a timed test, Alexander typed 742 words in 14 minutes. AssumingAlexander works at this rate for the next hour, which of the following bestapproximates the number of words he would type in that hour?A.53B.840C.3180D.44,520Correct answer: C4.A quilter wants to make the design shown above using the Golden Ratio.Specifically, he wants the ratio of the triangle heights A:B and B:C to eachequal 1.62. If the quilter makes the triangle height A=8 in, approximately how tall should he make triangle height C?A.3 inB. 4 inC. 5 inD.6 inCorrect answer: A5.Elena is conducting a study about the effects of toxins in the water on thehormones of fish. Elena surveys 350 male fish in a river and finds that 150 of the male fish have egg cells growing inside them. According to Elena's survey, what is the ratio of male fish with egg cells to male fish without egg cells in the river?A.3:4B.3:7C.4:5D.4:7Correct answer: A。

可汗学院数学题目-18Polynomialfactorsandgraphs

Polynomial factors and graphs1.P(x) =2x3−18xGiven the polynomial function P defined above, what are its zeros?A. {−9,−6,2,3}B. {−9,0,2}C. {−3,3}D. {−3,0,3}Correct Answer: D Difficult Level: 22. Which of the following functions could represent the graph BELOW in the xy-plane, where y=P(x)?A. P(x) =x2+6x+8B. P(x) =x3+6x2+8xC. P(x) =x2−6x+8D. P(x) =x3−6x2+8xCorrect Answer: B Difficult Level: 23. A polynomial has zeros at−9,2, and0. Which of the following could be the polynomial?A. x2−7x−18B.x3+7x2−18xC.x3+7x2−18xD.x3+6x2−25x+18Correct Answer: B Difficult Level: 24. (x−7) (x+5) (2x−3)Given the polynomial above, what are its zeros?A.｛-7, 5, -3｝B. {7, - 5, 3}C. {-7, 5, -3}2D. {7, -5, 3}2Correct Answer: D Difficult Level: 25. 2(x+55) (x−17)Given the polynomial above, what are its zeros?A. x=−55and x=17B. x=−55, x=−2, and x=17C. x=−17and x=55D. x=−17,x=2, and x=55Correct Answer: A Difficult Level: 26. G (z) = (z−1)5−(z−1)4 The polynomial function G is defined above. What is the product of the zeros of G? Fill in the blank:Correct Answer: 2 Difficult Level: 27. The polynomial function P has zeros at3and6. Which of the following could be the definition of P?A. P(x)=x2+9x+18B. P(x)=x2−9x+18C. P(x)=x2+3x+6D. P(x)=x2+6x+3Correct Answer: B Difficult Level: 28. Which of the following equations could represent the graph below in the xy-plane?I. y=(x−3)(x+3)II. y=(x−4)2III. y=(x+2)(x+7)A. ⅠonlyB. ⅡonlyC. I and ⅢonlyD. I,Ⅱand ⅢCorrect Answer: A Difficult Level: 29. h (t)=(t−8)1(t−4)2(t−2)3(t−1)4The polynomial function h is defined above. How many distinct zeros does h have? Correct Answer: 4 Difficult Level: 210.x3 +25x2+50x−1000The polynomial above has(x−5)and (x+10)as factors. What is the remaining factor?A. (x+2)B.(x-2)C. (x+2)D. (x-20)Correct Answer: C Difficult Level: 211.x3+7x2−36The polynomial above has zeros at −6 and 2. If the remaining zero is z, then what is the value of −z?Correct Answer: 3 Difficult Level: 312. The function p is a polynomial of t such that (t−10), (22−t),(t+10), and (20+t)are all factors of p(t). Which of the following could be the graph of y=p(t)in the ty-plane?A. B.C. D.Correct Answer: A Difficult Level: 313. Which of the following graphs appears to represent a polynomial function with a double zero?A. B.C. D.Correct Answer: D Difficult Level: 314.g(x)=x4−4x3+6x2−4x+1The function g is defined above. Given that all zeros of g are integers between−1 and 1inclusive, how many distinct zeros does g have?Correct Answer: 1 Difficult Level: 315.x2−ax+24If one of the zeros of the polynomial above is 8, what is the other zero?Correct Answer: 3 Difficult Level: 316. For a function g, the graph of y=g(x)y=g(x)is shown BELOW. When g(x)is divided by(x+10), the remainder is−20. Which of the following is closest to the remainder when g(x)is divided by(x−10)?A. -28B. -2C. 2D. 28Correct Answer: C Difficult Level: 317. The polynomial function f is defined as f(c)=(c−k)(c2−4c+4) where k is a constant. The value 2 is a zero of f. What is the remainder of f(c) when divided by (c−2)?Correct Answer: 0 Difficult Level: 318. Which of the following graphs in the xy-plane have−3and5as all of their distinct zeros for −6≤x≤6?I. II.III.A. I onlyB. I and II onlyC. II and III onlyD. I. II and III.Correct Answer: B Difficult Level: 319. Given some rational constant a, which polynomial equation could represent the graph BELOW in the xy-plane?A.y=a(x+1)2(x−5)B.y=a(x+1) (x−5)C.y=a(x+1) (2x+1)(x−5)D.y=a(x−1) (x+5)2Correct Answer: B Difficult Level: 320. The equation s=(t+3)2(t+2)(t+1)(t)(t−1) is graphed on the st-plane. What is the product of the t-intercepts of the graph?Correct Answer: 0 Difficult Level: 321. q(v)=(v−8)(v−5)(v−4)(v+5)(v+10)The function q is defined above. If the sum of the zeros of q is s, what is the value of s?Correct Answer: 2 Difficult Level: 322. (x−√3)2(x−√7)Given the polynomial above, what are its zeros?A.x=−√3and x=−√7B. x=√3and x=√7C.x=3and x=√7D.x=−3and x=−√7Correct Answer: B Difficult Level: 323. The graph of the polynomial equation y=α(t) is shown BELOW. Which of the following must be true?** leading coefficient is positive.** sum of the distinct t intercepts is negative.** constant coefficient is positive.** product of the distinct t intercepts is negative.Correct Answer: D Difficult Level: 324. The graph shown at left could represent which of the following equations?A. h=−(b−10)(b−20)(b+20)B. h=(b−10)(b−20)(b+20)C. h=−(b+10)(b−20)(b+20)D. h=(b+10)(b−20)(b+20)Correct Answer: A Difficult Level: 325.x3+7x2−36The polynomial above has zeros at −6 and 2. If the remaining zero is z, then what is the value of−z?Correct Answer: 3 Difficult Level: 326. The polynomial function f is defined as f(m)=(m3−m2−17m−15)(m+1). Whenf(m) is divided by (m+1), what is the remainder?Correct Answer: 0 Difficult Level: 427.p(n)=(n3−12n2+47n−60)(n−4)q(n)=(n+13)(n−4)The functions p and q are defined above. One of the functions has a zero at n=5. What is (p+q)(5)?Correct Answer: 18 Difficult Level: 428. Which of the following could be the equation corresponding to the graph BELOW?A.s=a⋅a⋅a⋅aB.s=(a−1)(a−1)C.s=a⋅a⋅a⋅a⋅aD.s=(a−1)(a−1)(a−1)Correct Answer: C Difficult Level: 429.p=(w−30)(w2+178w+7921)Given that −89 is a double zero of the polynomial equation above, which of the following could be the graph of the equation in the pw-plane?A. B.C. D.Correct Answer: D Difficult Level: 430.ℓ(x)=x4+36x2−10,000The polynomial function ℓ is defined above. What is the remainder of ℓ(x) when divided by (x+10)?Correct Answer: 3600 Difficult Level: 431.g (w)=(w+13)3(w+19)2The polynomial function g is defined above. When g (w) is divided by (w+16), the remainder is r. What is the value of ∣r∣?Correct Answer: 243 Difficult Level: 432. A function p p is defined as p(x) = (x−a)(x−15)(x−20)+15 where a is a constant. Given that p (7)=15, what is the value of a?Correct Answer: 7 Difficult Level: 433. A function w is a defined as w(x)=ax2+bx+c where a, b, and c are constants.If a=3 and w(3)=w(15)=0, then what is the absolute value of b?Correct Answer: 54 Difficult Level: 434. A function s is defined as s(x)=(x−4)(x−5)2. A function h is defined ash(x)=(x−a)s(x). For some constant a, (x−a)3 is a factor of h. What is s(a)?Correct Answer: 0 Difficult Level: 435.12x2+ax+2, what is the other zero?If one of the zeros of the above polynomial is 23Correct Answer: 1Difficult Level: 44** any polynomial function h, the polynomial function g is defined to be g(w)=(w−25 )(w−1)h(w). If h has zeros at 1 and 3 only, what is the sum of the distinct zeros of g?Correct Answer: 36 Difficult Level: 437. The polynomial function f f is defined as f(x)=(x−c1)(x−c2)(x−c3)⋯(x−cn)for some positive integer n. Each of the values c1,c2,c3,,,cn is a real number. The graph of y=f(x)is shown BELOW. Which of the following could be the value of n?A. 0B. 1C. 2D. 3Correct Answer: D Difficult Level: 438. The polynomial function f is defined as f(m)=(m3−m2−17m−15)(m+1). When f(m) is divided by (m+1), what is the remainder?Correct Answer: 0 Difficult Level: 4。

可汗学院微课程分析及启示——以比例（Ratio）为例

可汗学院微课程分析及启示——以比例（Ratio）为例
钱丽娟
【期刊名称】《中学数学研究（下半月）》
【年(卷),期】2015(000)006
【摘要】社会的发展,信息技术的进步,要求我们更新教育认识的同时,也为教育新模式的发展和实现提供了契机.以可汗学院为代表的,风靡全球的微型网络教学视频的出现激发了教育研究者对以微视频为载体运用于课堂的教学探索,滋生了一批新的名词,如翻转课堂,慕课,微课,微课程等.[1]在进一步学习可汗学院微课程后,笔者开始反思其与传统教学的反差,希望以其中——比例（Ratio）为例,通过分析比例及比例习题课中微课程的特点、结构,为我们反思教学提供启示.
【总页数】3页(P8-10)
【作者】钱丽娟
【作者单位】浙江省浙江师范大学,321004
【正文语种】中文
【中图分类】TP3
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SAT考试专题2024数学历年题目解析

SAT考试专题2024数学历年题目解析2024年的SAT考试将继续囊括数学科目，下面将对该年度的数学部分历年题目进行解析，帮助考生更好地准备SAT数学考试。

1. 第一题解析该题目是一道代数题，要求求解方程：3x + 5 = 20。

解题思路：将方程中的变量与常数项分离，得到：3x = 20 - 5。

计算得：3x = 15，再将等式两边同时除以3，得到：x = 5。

因此，方程的解为x = 5。

2. 第二题解析该题目是一道几何题，要求计算三角形的面积。

解题思路：已知三角形的底边长度为6，高为8。

直接使用三角形面积公式：面积 = 底边长度 ×高 ÷ 2。

代入已知的数值进行计算：面积 = 6 × 8 ÷ 2 = 24。

因此，该三角形的面积为24平方单位。

3. 第三题解析该题目是一道概率题，要求计算从一副标准扑克牌中随机抽取一张牌，该牌为红桃的概率。

解题思路：一副标准扑克牌中共有52张牌，其中有13张红桃牌。

因此，红桃牌的概率为：概率 = 红桃牌数目 ÷总牌数目。

代入已知数值进行计算：概率 = 13 ÷ 52 = 1 ÷ 4 = 0.25。

因此，从一副标准扑克牌中随机抽取一张牌，该牌为红桃的概率为0.25。

4. 第四题解析该题目是一道函数题，要求计算函数的值。

解题思路：已知函数 f(x) = 2x^2 + 3x + 1，需要计算当 x = 2 时的函数值。

将 x = 2 代入函数表达式中，得到：f(2) = 2 × 2^2 + 3 × 2 + 1。

计算得：f(2) = 8 + 6 + 1 = 15。

因此，当 x = 2 时，函数 f(x) 的值为15。

5. 第五题解析该题目是一道统计题，要求根据给定的数据计算平均数。

解题思路：已知一组数据为：10, 12, 15, 18, 20。

需要计算这组数据的平均数。

平均数的计算公式为：平均数 = 总和 ÷数据个数。

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LEVEL 31.Playing tennis Playing squashEnergyexpenditurerate 1,594kJℎ2,726kJℎOxygenconsumptionrate 77Lℎ130LℎFat burning rate 49 gℎrThe table above lists average expenditure rates of playing tennis vs. playing squash based on a body weight of 55 kilograms. The ratio of each value in the tennis column to the corresponding value in the squash column is about the same throughout the table. Which of the following is closest to the fat burning rate r of squash?A.27g ℎB.29g ℎC.83g ℎD.85g ℎCorrect answer: C2.Lea’s car travels an average of 30 miles per gallon of gas. If she spent $20.70 ongas for a 172.5 mile trip, what was the approximate cost of gas in dollars per gallon?A.$1.45 per gallonB.$3.40 per gallonC.$3.60 per gallonD.$5.75 per gallonCorrect answer: C3.Kavitha and Andrei are truck drivers for the Delightful Delivery Company,which services cities W, T, and Q.The distance from city W to city T is 48.5 miles.The distance from city T to city Q is 20 miles.The distance from city Q to city W is 36 miles.If Kavitha's truck and Andrei's truck travel at the same speed, and it takesAndrei 91 minutes to go from city W to city T, about how much time will it take Kavitha to travel from city W to city Q?A.19 minutesB.38 minutesC.68 minutesD.123 minutesCorrect answer: C4.Yasemin is using a sugar cookie recipe with a flour to sugar ratio of 11:6. Tomake one batch of cookies, 2.75 cups of flour are needed. If Yasemin would like to triple the recipe, which of the following best approximates how much sugar she will need?B.** cups5 cupsC.15 cupsCorrect answer: B5. A certain high school geometry class is made up of freshmen and sophomores.The ratio of freshmen to sophomores in that class is 3:4. If thereare 12 sophomores in the class, what is the total number of students in the class?A.9B.16C.21D.28Correct answer: C1.The musical interval between two sounds is called an "octave" if the ratio of thesounds' frequencies is 2:1. The following table shows the names of the musical intervals between two sounds based on the ratios of the two sounds' frequencies. Name of the musical interval Ratio of two sound frequenciesMajor third 4:5Perfect fourth 3:4Perfect fifth 2:3Major sixth 3:5If a sound is played with a frequency of 480 Hz, and a second sound is played with a frequency of 800 Hz, what is the name of the musical interval between the two sounds?A.Major thirdB.Perfect fourthC.Perfect fifthD.Major sixthCorrect answer: D2.The "maximum occupancy" of a room is the total number of people who can be ina room without causing a fire hazard. In a large room, a Fire Safety Code statesthat the maximum occupancy is 1 person for every 7 square feet (ft2). A college is hosting a concert in a hall that is 14,721 ft2, and 1,000 people are expected to attend. According to the Fire Safety Code, approximately how many more people can attend the concert without causing a fire hazard?A.900 peopleB.1,000 peopleC.1,100 peopleD.2,100 peopleCorrect answer: CA **pany makes cream-filled donuts using14cup (c) of dough and12tablespoon (tbsp) of cream per donut. **pany decides to change their recipe to use 3 times the amount of cream for their "New Triple-Stuffed Donuts!" If the **pany's new recipe uses the same amount of dough per donut, what is the ratio of dough to cream needed to make 12 triple-stuffed donuts?A.1 c:3 tbspB. 1 c:6 tbspC. 2 c:3 tbspD.1 c:2 tbspCorrect answer: B3. A marine aquarium has a small tank and a large tank, each containing only red andblue fish. In each tank, the ratio of red fish to blue fish is 3 to 4. The ratio of fish in the large tank to fish in the small tank is 46 to 5. What is the ratio of blue fish in the small tank to red fish in the large tank?A.15:184B.10:69C.69:10D.184:15Correct answer: B4.Two leading brands of paper towels are on sale. Brand A has 6 rolls, eachwith 56 sheets, for $4.29. Brand B has 8 rolls, each with 48 sheets, for $5.99.Which of the following best describes the relationship between the cost per sheet of the two brands?A.The two brands cost the same amount per sheet.B.Brand B costs $0.003 more per sheet than Brand A.C.Brand A costs $0.003 more per sheet than Brand B.D.Brand B costs $0.03 more than Brand A.Correct answer: BLEVEL 21.Zhang Lei spent $20.00 during his last outing at the bowling alley. This included aone time shoe rental fee of $3.50. He spent the rest of the money on bowling a number of games. If it took Zhang Lei 45 minutes to bowl each game and hespent 2 hours and 15 minutes bowling, how much did it cost per game?A.$3.00B.$5.50C.$6.67D.$7.34Correct answer: B Level：2Mr. Bumble delivers newspapers to **munity every week. His route takes 3 hours to complete. During the first 2 weeks of delivering newspapers, he traveled a totalof 100 blocks. Mr. Bumble delivers newspapers for 50 weeks each year. How many blocks does Mr. Bumble travel delivering newspapers in 1 year?A.1 blockB. 4 blocksC.50 blocksD.2,500 blocksCorrect answer: D2.During a timed test, Alexander typed 742 words in 14 minutes. AssumingAlexander works at this rate for the next hour, which of the following bestapproximates the number of words he would type in that hour?A.53B.840C.3180D.44,520Correct answer: C3.A quilter wants to make the design shown above using the Golden Ratio.Specifically, he wants the ratio of the triangle heights A:B and B:C to eachequal 1.62. If the quilter makes the triangle height A=8 in, approximately how tall should he make triangle height C?A.3 inB. 4 inC. 5 inD.6 inCorrect answer: A4.Elena is conducting a study about the effects of toxins in the water on thehormones of fish. Elena surveys 350 male fish in a river and finds that 150 of the male fish have egg cells growing inside them. According to Elena's survey, what is the ratio of male fish with egg cells to male fish without egg cells in the river?A.3:4B.3:7C.4:5D.4:7Correct answer: A。