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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)))

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

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) =?

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

Question Number 170426    Answers: 2   Comments: 0

Question Number 170416    Answers: 2   Comments: 0

∫(√(4−x^2 ))dx Mastermind

$$\int\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx} \\ $$$$ \\ $$$${Mastermind} \\ $$

Question Number 170415    Answers: 1   Comments: 0

Given x^y =e^(x−y) . find (dy/dx)? Mastermind

$$\mathrm{Given}\:\mathrm{x}^{\mathrm{y}} =\mathrm{e}^{\mathrm{x}−\mathrm{y}} .\:\mathrm{find}\:\frac{\mathrm{dy}}{\mathrm{dx}}? \\ $$$$ \\ $$$$\mathrm{Mastermind} \\ $$

Question Number 170413    Answers: 1   Comments: 0

Use Trapezoidal rule with ordinate 1, 1.5, 2, 2.5, 3 to compute ∫_1 ^3 ((√(x+1))/x)dx. Mastermind

$${Use}\:{Trapezoidal}\:{rule}\:{with}\:{ordinate} \\ $$$$\mathrm{1},\:\mathrm{1}.\mathrm{5},\:\mathrm{2},\:\mathrm{2}.\mathrm{5},\:\mathrm{3}\:{to}\:{compute} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{3}} \frac{\sqrt{{x}+\mathrm{1}}}{{x}}{dx}. \\ $$$$ \\ $$$${Mastermind} \\ $$

Question Number 170412    Answers: 2   Comments: 0

If part of the curve y=x+1 from x=1 and x=2 is rotated completely about the y−axis. find the volume of the solid form. Mastermind

$${If}\:{part}\:{of}\:{the}\:{curve}\:{y}={x}+\mathrm{1}\:{from}\:{x}=\mathrm{1} \\ $$$${and}\:{x}=\mathrm{2}\:{is}\:{rotated}\:{completely}\:{about} \\ $$$${the}\:{y}−{axis}.\:{find}\:{the}\:{volume}\:{of}\:{the} \\ $$$${solid}\:{form}. \\ $$$$ \\ $$$${Mastermind} \\ $$

Question Number 170411    Answers: 1   Comments: 0

The position of a body at time t (in seconds) is given by s=t^3 −4, what is the velocity and acceleration of the object at time t=5 ? Mastermind

$${The}\:{position}\:{of}\:{a}\:{body}\:{at}\:{time}\:{t}\:\left({in}\right. \\ $$$$\left.{seconds}\right)\:{is}\:{given}\:{by}\:{s}={t}^{\mathrm{3}} −\mathrm{4},\:{what}\:{is} \\ $$$${the}\:{velocity}\:{and}\:{acceleration}\:{of}\:{the} \\ $$$${object}\:{at}\:{time}\:{t}=\mathrm{5}\:? \\ $$$$ \\ $$$${Mastermind} \\ $$

Question Number 170410    Answers: 2   Comments: 0

Given y=xe^(−x) , show that (d^2 y/dx^2 )+2(dy/dx)+y=0 Mastermind

$${Given}\:{y}={xe}^{−{x}} ,\:{show}\:{that} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{2}\frac{{dy}}{{dx}}+{y}=\mathrm{0} \\ $$$$ \\ $$$${Mastermind} \\ $$

Question Number 170407    Answers: 0   Comments: 0

Question Number 170405    Answers: 0   Comments: 0

Question Number 170404    Answers: 0   Comments: 0

Question Number 170389    Answers: 1   Comments: 7

solve for x∈R x^3 +((3−x))^(1/3) −3=0

$${solve}\:{for}\:{x}\in{R} \\ $$$${x}^{\mathrm{3}} +\sqrt[{\mathrm{3}}]{\mathrm{3}−{x}}−\mathrm{3}=\mathrm{0} \\ $$

Question Number 170388    Answers: 0   Comments: 0

Question Number 170386    Answers: 1   Comments: 2

Find the volume of tetrahedron whose vertices are the points A(2, −1, −3), B(4, 1, 3), C(3, 2, −1) and D(1, 4, 2). Mastermind

$${Find}\:{the}\:{volume}\:{of}\:{tetrahedron}\:{whose} \\ $$$${vertices}\:{are}\:{the}\:{points}\:{A}\left(\mathrm{2},\:−\mathrm{1},\:−\mathrm{3}\right), \\ $$$${B}\left(\mathrm{4},\:\mathrm{1},\:\mathrm{3}\right),\:{C}\left(\mathrm{3},\:\mathrm{2},\:−\mathrm{1}\right)\:{and}\:{D}\left(\mathrm{1},\:\mathrm{4},\:\mathrm{2}\right). \\ $$$$ \\ $$$${Mastermind} \\ $$

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