Question and Answers Forum

All Questions   Topic List

AllQuestion and Answers: Page 1867

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

Question Number 20102    Answers: 2   Comments: 0

Solve the equation: (log _(sin x) cos x)^2 =1

$${Solve}\:{the}\:{equation}: \\ $$$$\left(\mathrm{log}\:_{\mathrm{sin}\:{x}} \mathrm{cos}\:{x}\right)^{\mathrm{2}} =\mathrm{1} \\ $$

Question Number 20091    Answers: 1   Comments: 0

Prove that Σ_(n=0) ^3 tan^2 (((2n + 1)π)/(16)) = 28.

$$\mathrm{Prove}\:\mathrm{that}\:\underset{{n}=\mathrm{0}} {\overset{\mathrm{3}} {\sum}}\mathrm{tan}^{\mathrm{2}} \:\frac{\left(\mathrm{2}{n}\:+\:\mathrm{1}\right)\pi}{\mathrm{16}}\:=\:\mathrm{28}. \\ $$

  Pg 1862      Pg 1863      Pg 1864      Pg 1865      Pg 1866      Pg 1867      Pg 1868      Pg 1869      Pg 1870      Pg 1871   

Terms of Service

Privacy Policy

Contact: info@tinkutara.com