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Question Number 20205 Answers: 1 Comments: 0

find the sin^(−1) diferentiation

$${find}\:{the}\:{sin}^{−\mathrm{1}} \:{diferentiation} \\ $$

Question Number 20204 Answers: 0 Comments: 0

Question Number 20203 Answers: 0 Comments: 1

Question Number 20201 Answers: 2 Comments: 0

^4 (√(49 − 20(√6))) =

$$\:^{\mathrm{4}} \sqrt{\mathrm{49}\:−\:\mathrm{20}\sqrt{\mathrm{6}}}\:= \\ $$

Question Number 20200 Answers: 1 Comments: 0

Question Number 20198 Answers: 1 Comments: 0

Find exact form of cos (tan^(−1) ((1/2)))

$$\mathrm{Find}\:\mathrm{exact}\:\mathrm{form}\:\mathrm{of} \\ $$$$\mathrm{cos}\:\left(\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$

Question Number 20195 Answers: 1 Comments: 0

Question Number 20194 Answers: 0 Comments: 5

A plane is drawn through the midpoint of a diagonal of a cube perpendicular to the diagonal. Determine the area of the figure resulting from the section of the cube cut by this plane if the edge of the cube is equal to a.

$${A}\:{plane}\:{is}\:{drawn}\:{through}\:{the}\: \\ $$$${midpoint}\:{of}\:{a}\:{diagonal}\:{of}\:{a}\:{cube} \\ $$$${perpendicular}\:{to}\:{the}\:{diagonal}. \\ $$$${Determine}\:{the}\:{area}\:{of}\:{the}\:{figure} \\ $$$${resulting}\:{from}\:{the}\:{section}\:{of}\:{the} \\ $$$${cube}\:{cut}\:{by}\:{this}\:{plane}\:{if}\:{the}\:{edge} \\ $$$${of}\:{the}\:{cube}\:{is}\:{equal}\:{to}\:\boldsymbol{{a}}. \\ $$

Question Number 20192 Answers: 0 Comments: 0

t_1 =3, t_n =3t_(n−1) +2 ....n>1

$$ \\ $$$${t}_{\mathrm{1}} =\mathrm{3},\:{t}_{{n}} =\mathrm{3}{t}_{{n}−\mathrm{1}} +\mathrm{2}\:\:\:\:\:....{n}>\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 20202 Answers: 0 Comments: 10

Is definite integral can have negative value? Because I think ∫_a ^b f(x) dx is total area below graph f(x) from x = a until x = b, so it can′t be negative

$$\mathrm{Is}\:\mathrm{definite}\:\mathrm{integral}\:\mathrm{can}\:\mathrm{have}\:\mathrm{negative}\:\mathrm{value}? \\ $$$$\mathrm{Because}\:\mathrm{I}\:\mathrm{think}\:\int_{{a}} ^{{b}} {f}\left({x}\right)\:{dx}\:\mathrm{is}\:\mathrm{total}\:\mathrm{area}\:\mathrm{below} \\ $$$$\mathrm{graph}\:{f}\left({x}\right)\:\mathrm{from}\:{x}\:=\:{a}\:\mathrm{until}\:{x}\:=\:{b},\:\mathrm{so}\:\mathrm{it}\:\mathrm{can}'\mathrm{t} \\ $$$$\mathrm{be}\:\mathrm{negative} \\ $$

Question Number 20182 Answers: 0 Comments: 0

Question Number 20187 Answers: 0 Comments: 1

t_n =(t_(n−1) /n^2 ), t_1 =3;t_2 ,t_3 ,(n≥2)

$${t}_{{n}} =\frac{{t}_{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} },\:{t}_{\mathrm{1}} =\mathrm{3};{t}_{\mathrm{2}} ,{t}_{\mathrm{3}} ,\left({n}\geqslant\mathrm{2}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 20177 Answers: 0 Comments: 0

But ans is (1/(108))

$${But}\:{ans}\:{is}\:\frac{\mathrm{1}}{\mathrm{108}} \\ $$

Question Number 20174 Answers: 2 Comments: 0

(0.1^− )^2 {1−9(0.16^− )^2 }

$$\left(\mathrm{0}.\overset{−} {\mathrm{1}}\right)^{\mathrm{2}} \left\{\mathrm{1}−\mathrm{9}\left(\mathrm{0}.\mathrm{1}\overset{−} {\mathrm{6}}\right)^{\mathrm{2}} \right\} \\ $$

Question Number 20167 Answers: 1 Comments: 0

please solve it integrate with respect to x ∫((5x−2)/(3x^2 +2x+1))

$${please}\:{solve}\:{it} \\ $$$${integrate}\:{with}\:{respect}\:{to}\:{x} \\ $$$$\int\frac{\mathrm{5}{x}−\mathrm{2}}{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}} \\ $$

Question Number 20166 Answers: 1 Comments: 0

∫cosec^2 xdx

$$\int\mathrm{cosec}\:^{\mathrm{2}} {xdx} \\ $$

Question Number 20164 Answers: 1 Comments: 0

∫(e^(tan^(−1) x) /(1+x^2 ))

$$\int\frac{{e}^{\mathrm{tan}^{−\mathrm{1}} {x}} }{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$

Question Number 20157 Answers: 1 Comments: 1

Question Number 20156 Answers: 1 Comments: 0

Solve: inverse laplace. L^(−1) ((s/(s^(2 ) + 6s + 25)))

$$\mathrm{Solve}:\:\mathrm{inverse}\:\mathrm{laplace}.\:\:\:\:\mathrm{L}^{−\mathrm{1}} \left(\frac{\mathrm{s}}{\mathrm{s}^{\mathrm{2}\:} +\:\mathrm{6s}\:+\:\mathrm{25}}\right) \\ $$

Question Number 20149 Answers: 2 Comments: 1

Question Number 20162 Answers: 1 Comments: 0

Compute the volume of a solid bounded by a surface with equation (x^2 +y^2 +z^2 )^2 =a^3 x .

$${Compute}\:{the}\:{volume}\:{of}\:{a}\:{solid} \\ $$$${bounded}\:{by}\:{a}\:{surface}\:{with}\:{equation} \\ $$$$\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\mathrm{2}} ={a}^{\mathrm{3}} {x}\:. \\ $$

Question Number 20138 Answers: 1 Comments: 0

lim_(x→0) ((1−cos ax)/(1−cos bx))

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:{ax}}{\mathrm{1}−\mathrm{cos}\:{bx}} \\ $$$$ \\ $$

Question Number 20118 Answers: 1 Comments: 0

The quadratic equations x^2 − 6x + a = 0 and x^2 − cx + 6 = 0 have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then, find the common root.

$$\mathrm{The}\:\mathrm{quadratic}\:\mathrm{equations}\:{x}^{\mathrm{2}} \:−\:\mathrm{6}{x}\:+\:{a}\:=\:\mathrm{0} \\ $$$$\mathrm{and}\:{x}^{\mathrm{2}} \:−\:{cx}\:+\:\mathrm{6}\:=\:\mathrm{0}\:\mathrm{have}\:\mathrm{one}\:\mathrm{root}\:\mathrm{in} \\ $$$$\mathrm{common}.\:\mathrm{The}\:\mathrm{other}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first} \\ $$$$\mathrm{and}\:\mathrm{second}\:\mathrm{equations}\:\mathrm{are}\:\mathrm{integers}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{ratio}\:\mathrm{4}\::\:\mathrm{3}.\:\mathrm{Then},\:\mathrm{find}\:\mathrm{the}\:\mathrm{common} \\ $$$$\mathrm{root}. \\ $$

Question Number 20116 Answers: 1 Comments: 0

If a and b (≠ 0) are the roots of the equation x^2 + ax + b = 0, then find the least value of x^2 + ax + b (x ∈ R).

$$\mathrm{If}\:{a}\:\mathrm{and}\:{b}\:\left(\neq\:\mathrm{0}\right)\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:{x}^{\mathrm{2}} \:+\:{ax}\:+\:{b}\:=\:\mathrm{0},\:\mathrm{then}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{least}\:\mathrm{value}\:\mathrm{of}\:{x}^{\mathrm{2}} \:+\:{ax}\:+\:{b}\:\left({x}\:\in\:{R}\right). \\ $$

Question Number 20115 Answers: 1 Comments: 0

The value of a for which the equation (1 − a^2 )x^2 + 2ax − 1 = 0 has roots belonging to (0, 1) is

$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:{a}\:\mathrm{for}\:\mathrm{which}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\left(\mathrm{1}\:−\:{a}^{\mathrm{2}} \right){x}^{\mathrm{2}} \:+\:\mathrm{2}{ax}\:−\:\mathrm{1}\:=\:\mathrm{0}\:\mathrm{has}\:\mathrm{roots} \\ $$$$\mathrm{belonging}\:\mathrm{to}\:\left(\mathrm{0},\:\mathrm{1}\right)\:\mathrm{is} \\ $$

Question Number 20110 Answers: 0 Comments: 1

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