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Question Number 37664 Answers: 1 Comments: 0

find ∫(1 + sinx)dx

$$\mathrm{find}\:\int\left(\mathrm{1}\:+\:{sinx}\right){dx} \\ $$

Question Number 37663 Answers: 3 Comments: 0

show that ((sin2A)/(1+cos2A)) = tanA.

$$\mathrm{show}\:\mathrm{that}\:\frac{{sin}\mathrm{2}{A}}{\mathrm{1}+{cos}\mathrm{2}{A}}\:=\:\mathrm{tanA}. \\ $$

Question Number 37662 Answers: 1 Comments: 0

Given that y=x^2 cosx, find (dy/(dx )), simplifying your answer as far as posible

$$\:\mathrm{Given}\:\mathrm{that}\:{y}={x}^{\mathrm{2}} {cosx}, \\ $$$$\mathrm{find}\:\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}\:},\:\mathrm{simplifying}\:\mathrm{your}\:\mathrm{answer} \\ $$$$\mathrm{as}\:\mathrm{far}\:\mathrm{as}\:\mathrm{posible} \\ $$

Question Number 37661 Answers: 1 Comments: 0

Given the lines l_1 : r= −5i + 2j + s(3i−j) l_2 : r= −2i+j + t(2i+j) Find the cosine of the angle between l_1 and l_2 .

$$\mathrm{Given}\:\mathrm{the}\:\mathrm{lines}\: \\ $$$${l}_{\mathrm{1}} :\:\mathrm{r}=\:−\mathrm{5}{i}\:+\:\mathrm{2}{j}\:+\:{s}\left(\mathrm{3}{i}−{j}\right) \\ $$$${l}_{\mathrm{2}} :\:{r}=\:−\mathrm{2}{i}+{j}\:+\:{t}\left(\mathrm{2}{i}+{j}\right) \\ $$$${F}\mathrm{ind}\:\mathrm{the}\:\mathrm{cosine}\:\mathrm{of}\:\mathrm{the}\:\mathrm{angle} \\ $$$$\mathrm{between}\:{l}_{\mathrm{1}} \:{and}\:{l}_{\mathrm{2}} . \\ $$

Question Number 37645 Answers: 1 Comments: 0

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Question Number 37249 Answers: 0 Comments: 0

a triangle with vertices A(2,1),B(6,1) and (3,3) is transformed by (((2 0)),((0 2)) ) a)find the image A′B′C′ after this transformstion b) state the type of transformation

$${a}\:{triangle}\:{with}\:{vertices}\: \\ $$$$\:{A}\left(\mathrm{2},\mathrm{1}\right),{B}\left(\mathrm{6},\mathrm{1}\right)\:{and}\:\left(\mathrm{3},\mathrm{3}\right)\:{is}\:{transformed} \\ $$$${by}\:\begin{pmatrix}{\mathrm{2}\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\end{pmatrix}\: \\ $$$$\left.{a}\right){find}\:{the}\:{image}\:{A}'{B}'{C}'\:{after}\: \\ $$$${this}\:{transformstion} \\ $$$$\left.{b}\right)\:{state}\:{the}\:{type}\:{of}\:{transformation} \\ $$

Question Number 37218 Answers: 1 Comments: 0

Proove that a) ((1−cosA + cos B − cos(A+B))/(1+cos A − cosB−cos(A+B)))= tan(A/2).cot (B/2) b) cosα cos(60−α)cos(60+α)= (1/4)cos3α

$$\:{Proove}\:{that} \\ $$$$\left.{a}\right)\:\:\frac{\mathrm{1}−{cosA}\:+\:{cos}\:{B}\:−\:{cos}\left({A}+{B}\right)}{\mathrm{1}+{cos}\:{A}\:−\:{cosB}−{cos}\left({A}+{B}\right)}=\:{tan}\frac{{A}}{\mathrm{2}}.{cot}\:\frac{{B}}{\mathrm{2}} \\ $$$$\left.{b}\right)\:{cos}\alpha\:{cos}\left(\mathrm{60}−\alpha\right){cos}\left(\mathrm{60}+\alpha\right)=\:\frac{\mathrm{1}}{\mathrm{4}}{cos}\mathrm{3}\alpha \\ $$$$ \\ $$

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Question Number 37123 Answers: 0 Comments: 1

draw ΔAB^ C and its image ΔA′B^′ C^′ after a reflection in line y=x if A(0,3),B(3,0),C(3,2).what is the line of symmetry of the two figures?

$${draw}\:\Delta{A}\overset{} {{B}C}\:{and}\:{its}\:{image}\:\:\Delta{A}'{B}^{'} {C}^{'} \: \\ $$$${after}\:{a}\:{reflection}\:{in}\:{line}\:{y}={x} \\ $$$${if}\:{A}\left(\mathrm{0},\mathrm{3}\right),{B}\left(\mathrm{3},\mathrm{0}\right),{C}\left(\mathrm{3},\mathrm{2}\right).{what}\:{is}\:{the} \\ $$$${line}\:{of}\:{symmetry}\:\:{of}\:{the}\:{two}\:{figures}? \\ $$

Question Number 37122 Answers: 0 Comments: 0

draw ΔAB^ C and its image ΔA′B^′ C^′ after a reflection in line y=x if A(0,3),B(3,0),C(3,2).what is the line of symmetry of the two figures?

Question Number 37089 Answers: 1 Comments: 10

Question Number 37084 Answers: 0 Comments: 2

solve using matrix method x − y= 4 2x − 3y= 5

$$\:\:{solve}\:{using}\:{matrix}\:{method} \\ $$$$\:\:\:\:\:\:{x}\:−\:{y}=\:\mathrm{4} \\ $$$$\:\:\:\:\:\mathrm{2}{x}\:−\:\mathrm{3}{y}=\:\mathrm{5} \\ $$

Question Number 37081 Answers: 0 Comments: 1

Question Number 37037 Answers: 4 Comments: 0

If ((a+b−c)/(a+b)) = ((b+c−a)/(b+c)) =((c+a−b)/(c+a)) and a+b+c ≠ 0 then which of the following is true 1) a=−b=c 2)−a=−b=c 3) a=b=c 4) a=b≠c

$${If}\:\frac{{a}+{b}−{c}}{{a}+{b}}\:=\:\frac{{b}+{c}−{a}}{{b}+{c}}\:=\frac{{c}+{a}−{b}}{{c}+{a}}\:{and}\: \\ $$$$\mathrm{a}+\mathrm{b}+\mathrm{c}\:\neq\:\mathrm{0}\:{then}\:{which}\:{of}\:{the}\:{following} \\ $$$$\left.{i}\left.{s}\:{true}\:\mathrm{1}\right)\:{a}=−{b}={c}\:\mathrm{2}\right)−{a}=−{b}={c}\: \\ $$$$\left.\mathrm{3}\left.\right)\:\mathrm{a}=\mathrm{b}=\mathrm{c}\:\mathrm{4}\right)\:\mathrm{a}=\mathrm{b}\neq{c} \\ $$

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