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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

$$ \\ $$

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)))

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

Question Number 170448    Answers: 0   Comments: 1

tan90°=

$$\mathrm{tan90}°= \\ $$

Question Number 170445    Answers: 1   Comments: 0

Solve ((sin x)/(sin (x+5°))) = (1/(2cos 55°))

$$\:\:\:{Solve}\:\:\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:\left({x}+\mathrm{5}°\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2cos}\:\mathrm{55}°} \\ $$

Question Number 170443    Answers: 0   Comments: 0

A body of weight 6 newtons is plased on a rough horizontal plane whose coefficient of friction is ((√3)/3) the friction force ε=

$${A}\:{body}\:{of}\:{weight}\:\mathrm{6}\:{newtons}\:{is}\:{plased}\:{on}\:{a}\:{rough}\:{horizontal}\:{plane}\:{whose}\:{coefficient}\:{of}\:{friction}\:{is}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:{the}\:{friction}\:{force}\:\epsilon= \\ $$

Question Number 170435    Answers: 1   Comments: 0

A line is formed by joining the points A(7,0) and B(0,2). Obtain the equation of the straight line joining AC such that the x−axis bisects the angle BAC.

$$\boldsymbol{{A}}\:\boldsymbol{{line}}\:\boldsymbol{{is}}\:\boldsymbol{{formed}}\:\boldsymbol{{by}}\:\boldsymbol{{joining}}\:\boldsymbol{{the}} \\ $$$$\boldsymbol{{points}}\:\boldsymbol{{A}}\left(\mathrm{7},\mathrm{0}\right)\:\boldsymbol{{and}}\:\boldsymbol{{B}}\left(\mathrm{0},\mathrm{2}\right).\:\boldsymbol{{Obtain}} \\ $$$$\boldsymbol{{the}}\:\boldsymbol{{equation}}\:\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{straight}}\:\boldsymbol{{line}} \\ $$$$\boldsymbol{{joining}}\:\boldsymbol{{AC}}\:\boldsymbol{{such}}\:\boldsymbol{{that}}\:\boldsymbol{{the}}\:\boldsymbol{{x}}−\boldsymbol{{axis}} \\ $$$$\boldsymbol{{bisects}}\:\boldsymbol{{the}}\:\boldsymbol{{angle}}\:\boldsymbol{{BAC}}. \\ $$

Question Number 170427    Answers: 1   Comments: 0

14^b =7^(b−a) 49^(a/b) =?

$$\mathrm{14}^{{b}} =\mathrm{7}^{{b}−{a}} \\ $$$$\mathrm{49}^{\frac{{a}}{{b}}} =? \\ $$

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