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AllQuestion and Answers: Page 492

Question Number 170511 Answers: 1 Comments: 1

what does relative charge mean?

$${what}\:{does}\:{relative}\:{charge}\:{mean}? \\ $$

Question Number 170522 Answers: 1 Comments: 0

Question Number 170503 Answers: 0 Comments: 1

tan90°=?

$$\mathrm{tan90}°=? \\ $$

Question Number 170502 Answers: 2 Comments: 0

Question Number 170501 Answers: 1 Comments: 0

solve this: ∫∫_D x^2 e^(xy) dxdy D:{(x.y)∈R^2 /0≤x≤1. 0≤y≤2} ∫∫_D ((ydxdy)/((1+x^2 +y^2 )^(3/2) )). 0≤x≤1.0≤y≤1.

$$\:\:\:\:\:\:\:\:\:\:{solve}\:{this}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\int_{{D}} {x}^{\mathrm{2}} {e}^{{xy}} {dxdy} \\ $$$${D}:\left\{\left({x}.{y}\right)\in{R}^{\mathrm{2}} \:/\mathrm{0}\leqslant{x}\leqslant\mathrm{1}.\:\:\mathrm{0}\leqslant{y}\leqslant\mathrm{2}\right\} \\ $$$$\:\:\:\:\:\:\int\underset{{D}} {\int}\frac{{ydxdy}}{\left(\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }.\:\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}.\mathrm{0}\leqslant{y}\leqslant\mathrm{1}. \\ $$

Question Number 170497 Answers: 0 Comments: 0

Given that R is at the point with positive vector 4−2j when t=2 Find the position vector of R when t=3

$${Given}\:{that}\:{R}\:{is}\:{at}\:{the}\:{point}\:{with} \\ $$$${positive}\:{vector}\:\mathrm{4}−\mathrm{2}{j}\:{when}\:{t}=\mathrm{2} \\ $$$${Find}\:{the}\:{position}\:{vector}\:{of}\:{R} \\ $$$$\:{when}\:{t}=\mathrm{3} \\ $$

Question Number 170494 Answers: 0 Comments: 0

Question Number 170495 Answers: 1 Comments: 0

Question Number 170492 Answers: 0 Comments: 0

Question Number 170491 Answers: 1 Comments: 0

1) Help 5^x −3^x =9

$$\left.\mathrm{1}\right)\:{Help} \\ $$$$\mathrm{5}^{{x}} −\mathrm{3}^{{x}} =\mathrm{9}\: \\ $$

Question Number 170488 Answers: 1 Comments: 0

Question Number 170486 Answers: 0 Comments: 0

Question Number 170485 Answers: 2 Comments: 0

Question Number 170480 Answers: 0 Comments: 3

Question Number 170479 Answers: 0 Comments: 0

Question Number 170478 Answers: 2 Comments: 0

(( 9 + 4(√5) ))^(1/3) + (( 9 − 4(√5) ))^(1/3) = ?

$$\sqrt[{\mathrm{3}}]{\:\mathrm{9}\:+\:\mathrm{4}\sqrt{\mathrm{5}}\:}\:+\:\sqrt[{\mathrm{3}}]{\:\mathrm{9}\:−\:\mathrm{4}\sqrt{\mathrm{5}}\:}\:=\:? \\ $$

Question Number 170475 Answers: 0 Comments: 0

An easy question of Measure theory: Prove that : μ ( Q )=0 ■m.n

$$ \\ $$$$\:\:\:\:\:\:\:{An}\:{easy}\:{question}\:{of}\:{Measure}\:{theory}: \\ $$$$ \\ $$$$\:\:\:\:\:{Prove}\:{that}\::\:\:\:\:\:\:\:\:\:\:\:\:\:\mu\:\left(\:\mathrm{Q}\:\right)=\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare{m}.{n} \\ $$$$ \\ $$

Question Number 170474 Answers: 0 Comments: 0

∫ (√(1+n^2 x^(2n−2) )) dx

$$\int\:\sqrt{\mathrm{1}+{n}^{\mathrm{2}} {x}^{\mathrm{2}{n}−\mathrm{2}} }\:{dx} \\ $$

Question Number 170473 Answers: 0 Comments: 0

Question Number 170472 Answers: 0 Comments: 0

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Question Number 170468 Answers: 1 Comments: 0

Given that log_4 (y−1)+log_4 ((x/y))=m and log_2 (y+1)−log_2 x=m−1, show that y^2 =1−8^m

$$\:\mathrm{G}{iven}\:{that}\:{log}_{\mathrm{4}} \left({y}−\mathrm{1}\right)+{log}_{\mathrm{4}} \left(\frac{{x}}{{y}}\right)={m} \\ $$$$\:{and}\:{log}_{\mathrm{2}} \left({y}+\mathrm{1}\right)−{log}_{\mathrm{2}} {x}={m}−\mathrm{1}, \\ $$$$\:{show}\:{that}\:{y}^{\mathrm{2}} =\mathrm{1}−\mathrm{8}^{{m}} \\ $$

Question Number 170463 Answers: 0 Comments: 1

Question Number 170461 Answers: 1 Comments: 1

find tbe domain and range of the following functions y=4x/x^2 −4

$${find}\:{tbe}\:{domain}\:{and}\:{range}\:{of}\:{the}\:{following}\:{functions} \\ $$$${y}=\mathrm{4}{x}/{x}^{\mathrm{2}} −\mathrm{4} \\ $$

Question Number 170459 Answers: 0 Comments: 0

Factorize 10x^2 y^2 +13x^2 y−3x^2 y−3x^2

$${Factorize}\:\mathrm{10}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{13}{x}^{\mathrm{2}} {y}−\mathrm{3}{x}^{\mathrm{2}} {y}−\mathrm{3}{x}^{\mathrm{2}} \\ $$

Question Number 170458 Answers: 1 Comments: 0

If the line y=mx+c is a tangent of a circle x^2 +y^2 =r^2 . Show that, c^2 =r^2 (1+m^2 )

$${If}\:{the}\:{line}\:{y}={mx}+{c}\:{is}\:{a}\:{tangent}\:{of} \\ $$$${a}\:{circle}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} .\:{Show}\:{that},\: \\ $$$$\:{c}^{\mathrm{2}} ={r}^{\mathrm{2}} \left(\mathrm{1}+{m}^{\mathrm{2}} \right) \\ $$

Question Number 170449 Answers: 0 Comments: 0

Solve ((sin 12°)/(sin 24° sin x)) = ((sin 72°)/(sin (36°+x)))

$$\:\:\:{Solve}\:\frac{\mathrm{sin}\:\mathrm{12}°}{\mathrm{sin}\:\mathrm{24}°\:\mathrm{sin}\:{x}}\:=\:\frac{\mathrm{sin}\:\mathrm{72}°}{\mathrm{sin}\:\left(\mathrm{36}°+{x}\right)} \\ $$

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