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Question Number 8528 Answers: 2 Comments: 0

A(2,b) translation T_1 = (((−3)),(( b)) ) followed by translation T_2 = ((a),(4) ) A′=(b−1,a−3) determine the value of a+b=...?

$${A}\left(\mathrm{2},{b}\right)\:{translation}\:{T}_{\mathrm{1}} =\begin{pmatrix}{−\mathrm{3}}\\{\:\:\:{b}}\end{pmatrix} \\ $$$${followed}\:{by}\:{translation}\:{T}_{\mathrm{2}} =\begin{pmatrix}{{a}}\\{\mathrm{4}}\end{pmatrix} \\ $$$${A}'=\left({b}−\mathrm{1},{a}−\mathrm{3}\right) \\ $$$${determine}\:{the}\:{value}\:{of}\:{a}+{b}=...? \\ $$

Question Number 8527 Answers: 1 Comments: 0

(√(6−(√(32)))) =...?

$$\sqrt{\mathrm{6}−\sqrt{\mathrm{32}}}\:=...? \\ $$

Question Number 8526 Answers: 1 Comments: 0

(√(8+2(√5))) = ...?

$$\sqrt{\mathrm{8}+\mathrm{2}\sqrt{\mathrm{5}}}\:=\:...? \\ $$

Question Number 8517 Answers: 0 Comments: 2

Question Number 8516 Answers: 2 Comments: 1

Question Number 8515 Answers: 0 Comments: 3

solve y=px+p^3 , p=(dy/dx)

$$\mathrm{solve}\:\:\mathrm{y}=\mathrm{px}+\mathrm{p}^{\mathrm{3}} ,\:\mathrm{p}=\frac{\mathrm{dy}}{\mathrm{dx}} \\ $$

Question Number 8514 Answers: 0 Comments: 1

show that∫_0 ^1 (x^2 /(√(1−x^4 )))dx=(1/4)β((3/4),(1/2))

$$\mathrm{show}\:\mathrm{that}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}^{\mathrm{2}} }{\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{4}} }}\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{4}}\beta\left(\frac{\mathrm{3}}{\mathrm{4}},\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$

Question Number 8510 Answers: 0 Comments: 0

Question Number 8503 Answers: 1 Comments: 2

Q. 1 sinA+sinB+sinC = 4cos(A/2) cos (B/2) cos (C/2) . Q.2 cosA cosB − cosC = 4cos(A/2) cos(B/2) cos(C/2) −1 Q.3 ((sin2A+sin2B+sin2C)/(sinA+sinB+sinC)) =8sin (A/2) sin(B/2) sin(C/2)

$${Q}.\:\mathrm{1}\:\:{sinA}+{sinB}+{sinC}\:=\:\mathrm{4}{cos}\frac{{A}}{\mathrm{2}}\:{cos} \\ $$$$\frac{{B}}{\mathrm{2}}\:{cos}\:\frac{{C}}{\mathrm{2}}\:. \\ $$$${Q}.\mathrm{2}\:\:{cosA}\:{cosB}\:−\:{cosC}\:=\:\mathrm{4}{cos}\frac{{A}}{\mathrm{2}}\:{cos}\frac{{B}}{\mathrm{2}} \\ $$$${cos}\frac{{C}}{\mathrm{2}}\:−\mathrm{1} \\ $$$$ \\ $$$${Q}.\mathrm{3}\:\:\frac{{sin}\mathrm{2}{A}+{sin}\mathrm{2}{B}+{sin}\mathrm{2}{C}}{{sinA}+{sinB}+{sinC}}\:=\mathrm{8}{sin}\:\frac{{A}}{\mathrm{2}} \\ $$$${sin}\frac{{B}}{\mathrm{2}}\:{sin}\frac{{C}}{\mathrm{2}} \\ $$$$ \\ $$

Question Number 8498 Answers: 0 Comments: 1

Question Number 8490 Answers: 1 Comments: 2

show thats true: ∫_(−∞) ^(+∞) e^(−x^2 ) = (√π)

$${show}\:{thats}\:{true}: \\ $$$$\int_{−\infty} ^{+\infty} {e}^{−{x}^{\mathrm{2}} } =\:\sqrt{\pi} \\ $$

Question Number 8488 Answers: 0 Comments: 7

find y_n if y=cos 2x

$${find}\:{y}_{{n}} \:{if}\:{y}=\mathrm{cos}\:\mathrm{2}{x} \\ $$

Question Number 8486 Answers: 1 Comments: 0

∫_(−∞) ^∞ e^(−2∣x∣^ ) (1−cos 2αx)dx=???? solve this .......

$$\underset{−\infty} {\overset{\infty} {\int}}{e}^{−\mathrm{2}\mid{x}\overset{} {\mid}\:} \left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\alpha{x}\right){dx}=???? \\ $$$${solve}\:{this}\:....... \\ $$$$ \\ $$$$ \\ $$

Question Number 8471 Answers: 1 Comments: 2

∫_(−∞) ^∞ e^(−2∣x∣dx_ ) ?

$$\underset{−\infty} {\overset{\infty} {\int}}{e}^{−\mathrm{2}\mid{x}\mid{d}\underset{} {{x}}} \:\:\:\:\:\:?\:\: \\ $$

Question Number 8468 Answers: 1 Comments: 0

Prove or disprove that: (2k+1)^n ∈O ∀k,n∈Z

$$\mathrm{Prove}\:\mathrm{or}\:\mathrm{disprove}\:\mathrm{that}: \\ $$$$\left(\mathrm{2}{k}+\mathrm{1}\right)^{{n}} \in\mathbb{O}\:\:\:\:\:\:\forall{k},{n}\in\mathbb{Z} \\ $$

Question Number 8466 Answers: 0 Comments: 1

t=n^3 −n^2 t∈E Give a general form solving for n i.e n^3 −n^2 =2k n=?

$${t}={n}^{\mathrm{3}} −{n}^{\mathrm{2}} \\ $$$${t}\in\mathbb{E} \\ $$$$\mathrm{Give}\:\mathrm{a}\:\mathrm{general}\:\mathrm{form}\:\mathrm{solving}\:\mathrm{for}\:{n} \\ $$$${i}.{e} \\ $$$${n}^{\mathrm{3}} −{n}^{\mathrm{2}} =\mathrm{2}{k} \\ $$$${n}=? \\ $$

Question Number 8465 Answers: 1 Comments: 0

Question Number 8452 Answers: 1 Comments: 3

Question Number 8421 Answers: 0 Comments: 1

if x=((a(1−r^n ))/(1−r)); make r the subject of the formulae

$${if}\:{x}=\frac{{a}\left(\mathrm{1}−{r}^{{n}} \right)}{\mathrm{1}−{r}};\:{make}\:{r}\:{the}\: \\ $$$${subject}\:{of}\:{the}\:{formulae} \\ $$

Question Number 8419 Answers: 4 Comments: 0

1) diket f(x)=2x−13, g^(−1) (x)=((x+4)/5) dan h^(−1) (x)=5x+7 nilai (f o ( g o h ))^(−1) (3)=...? 2)diket f(x)^(−1) =4x+5, g(x)=((x+4)/5) dan h^(−1) (x)=x−7 nilai ( f o g o h )^(−1) (−2)=....?? 3) jika diket invers dari fungsi f adalah f^(−1) (x)=3x^2 +2 dan invers dari fungsi g adalah g^(−1) (x)=(√(x+1)) × ≥−1 maka ( g o f )^(−1) (x)=....? 4) jika f(x)=2x−3 dan g(x)=(1/(3x+1)) utk x≠−(1/3) maka (g o f)^(−1) (x−(1/2))=....? sebelumnya saya ucapkan banyak terima kasih...

Question Number 8416 Answers: 0 Comments: 7

Solve equation 1. x^2 +y^2 =x+y+8 (x;y be positive) 2. x^2 −2(√x)+1=0

$$\boldsymbol{{Solve}}\:\boldsymbol{{equation}}\: \\ $$$$\:\mathrm{1}.\:\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} =\boldsymbol{{x}}+\boldsymbol{{y}}+\mathrm{8}\:\:\:\:\:\:\:\left({x};{y}\:{be}\:{positive}\right) \\ $$$$\:\mathrm{2}.\:\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{2}\sqrt{\boldsymbol{{x}}}+\mathrm{1}=\mathrm{0} \\ $$$$ \\ $$

Question Number 8427 Answers: 1 Comments: 0

In a circle PQRST center O, PQRS is a cyclic quadrilateral and T is on the circle. QS is a diameter and angle QOR is 86° if PTQ is 28° . Find the angles of the quadrilateral PQRS.

$$\mathrm{In}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{PQRST}\:\mathrm{center}\:\mathrm{O},\:\mathrm{PQRS}\:\mathrm{is}\:\mathrm{a}\: \\ $$$$\mathrm{cyclic}\:\mathrm{quadrilateral}\:\mathrm{and}\:\mathrm{T}\:\mathrm{is}\:\mathrm{on}\:\mathrm{the}\:\mathrm{circle}. \\ $$$$\mathrm{QS}\:\mathrm{is}\:\mathrm{a}\:\mathrm{diameter}\:\mathrm{and}\:\mathrm{angle}\:\mathrm{QOR}\:\mathrm{is}\:\mathrm{86}° \\ $$$$\mathrm{if}\:\mathrm{PTQ}\:\mathrm{is}\:\mathrm{28}°\:.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{angles}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{quadrilateral}\:\mathrm{PQRS}. \\ $$

Question Number 8411 Answers: 1 Comments: 0

ditentukan fungsi f:R→R, g:R→R dan h:R→R dg f(x)=(1/(x+4 )) , g(x)=3x dan h(x)=×−1 rumus ( h o g o f )^(−1) (1−x)=....?

$$\mathrm{ditentukan}\:\mathrm{fungsi}\:\mathrm{f}:\mathrm{R}\rightarrow\mathrm{R},\:\mathrm{g}:\mathrm{R}\rightarrow\mathrm{R}\:\mathrm{dan}\:\mathrm{h}:\mathrm{R}\rightarrow\mathrm{R}\: \\ $$$$\mathrm{dg}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{x}+\mathrm{4}\:}\:,\:\mathrm{g}\left(\mathrm{x}\right)=\mathrm{3x}\:\mathrm{dan}\:\mathrm{h}\left(\mathrm{x}\right)=×−\mathrm{1} \\ $$$$\mathrm{rumus}\:\left(\:\mathrm{h}\:\mathrm{o}\:\mathrm{g}\:\mathrm{o}\:\mathrm{f}\:\right)^{−\mathrm{1}} \left(\mathrm{1}−\mathrm{x}\right)=....? \\ $$$$ \\ $$$$ \\ $$

Question Number 8410 Answers: 1 Comments: 0

Question Number 8407 Answers: 0 Comments: 1

Question Number 8403 Answers: 3 Comments: 0

(Q.1) cos 4A=1−8cos^2 A + 8cos^2 A (Q.2) ((sec8 A −1)/(sec4 A −1)) = ((tan 8A)/(tan 2A)) (Q.3) tanA+tan(60^0 +A)+tan(120^0 +4) = 3tan 3A (Q.4) sinA sin (60^0 −A) sin(60^0 +A) = (1/4)sin3A

$$\left({Q}.\mathrm{1}\right)\:{cos}\:\mathrm{4}{A}=\mathrm{1}−\mathrm{8}{cos}^{\mathrm{2}} {A}\:+\:\mathrm{8}{cos}^{\mathrm{2}} \:{A} \\ $$$$\left({Q}.\mathrm{2}\right)\:\:\frac{{sec}\mathrm{8}\:{A}\:−\mathrm{1}}{{sec}\mathrm{4}\:{A}\:−\mathrm{1}}\:=\:\frac{{tan}\:\mathrm{8}{A}}{{tan}\:\mathrm{2}{A}} \\ $$$$\left({Q}.\mathrm{3}\right)\:\:\:{tanA}+{tan}\left(\mathrm{60}^{\mathrm{0}} +{A}\right)+{tan}\left(\mathrm{120}^{\mathrm{0}} \right. \\ $$$$\left.+\mathrm{4}\right)\:=\:\mathrm{3}{tan}\:\mathrm{3}{A}\: \\ $$$$\left({Q}.\mathrm{4}\right)\:\:\:{sinA}\:{sin}\:\left(\mathrm{60}^{\mathrm{0}} −{A}\right)\:\:{sin}\left(\mathrm{60}^{\mathrm{0}} +{A}\right)\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}{sin}\mathrm{3}{A} \\ $$

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