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Question Number 18907 Answers: 0 Comments: 3

Why Does A^→ .B^→ =ABCosθ?

$${Why}\:{Does}\: \\ $$$$\overset{\rightarrow} {{A}}.\overset{\rightarrow} {{B}}={ABCos}\theta?\: \\ $$

Question Number 18906 Answers: 1 Comments: 0

calculate (dy/dx) where y=cos^(−1) ((a+b cosx)/(b+a cosx)) (b>a)

$$\mathrm{calculate}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:\:\:\mathrm{where}\: \\ $$$$\mathrm{y}=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{a}+\mathrm{b}\:\mathrm{cosx}}{\mathrm{b}+\mathrm{a}\:\mathrm{cosx}}\:\left(\mathrm{b}>\mathrm{a}\right) \\ $$

Question Number 18905 Answers: 0 Comments: 0

prove that the differentiation of(((√(1+x^2 ))−(√(1−x^2 )))/((√(1+x^2 ))+(√(1−x^2 )))) with respect to (√(1−x^4 )) is ((√(1−x^4 ))/x^6 )

$$\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{differentiation} \\ $$$$\mathrm{of}\frac{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }−\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }+\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\:\:\mathrm{with}\:\mathrm{respect} \\ $$$$\mathrm{to}\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{4}} }\:\:\mathrm{is}\:\:\frac{\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{4}} }}{\mathrm{x}^{\mathrm{6}} } \\ $$

Question Number 18904 Answers: 1 Comments: 0

Solve the triangle in which a=((√3)+1), b=((√3)−1) and ∠C=60°.

$${Solve}\:{the}\:{triangle}\:{in}\:{which}\:{a}=\left(\sqrt{\mathrm{3}}+\mathrm{1}\right),\: \\ $$$${b}=\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\:{and}\:\angle{C}=\mathrm{60}°. \\ $$$$ \\ $$

Question Number 18893 Answers: 1 Comments: 0

Find all values of x, y and k for which the system of equations sin x cos 2y = k^4 − 2k^2 + 2 cos x sin 2y = k + 1 has a solution.

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{values}\:\mathrm{of}\:{x},\:{y}\:\mathrm{and}\:{k}\:\mathrm{for}\:\mathrm{which} \\ $$$$\mathrm{the}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations} \\ $$$$\mathrm{sin}\:{x}\:\mathrm{cos}\:\mathrm{2}{y}\:=\:{k}^{\mathrm{4}} \:−\:\mathrm{2}{k}^{\mathrm{2}} \:+\:\mathrm{2} \\ $$$$\mathrm{cos}\:{x}\:\mathrm{sin}\:\mathrm{2}{y}\:=\:{k}\:+\:\mathrm{1} \\ $$$$\mathrm{has}\:\mathrm{a}\:\mathrm{solution}. \\ $$

Question Number 18892 Answers: 1 Comments: 0

If the equation sin6x + cos4x = −2 have a family of nonnegative solutions x_k ′s, where 0 ≤ x_1 < x_2 < x_3 < .... < x_k < x_(k+1) ....., then the value of (1/π)Σ_(k=1) ^(1000) ∣x_(k+1) − x_k ∣ is

$$\mathrm{If}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{sin6}{x}\:+\:\mathrm{cos4}{x}\:=\:−\mathrm{2}\:\mathrm{have} \\ $$$$\mathrm{a}\:\mathrm{family}\:\mathrm{of}\:\mathrm{nonnegative}\:\mathrm{solutions}\:{x}_{{k}} '\mathrm{s}, \\ $$$$\mathrm{where}\:\mathrm{0}\:\leqslant\:{x}_{\mathrm{1}} \:<\:{x}_{\mathrm{2}} \:<\:{x}_{\mathrm{3}} \:<\:....\:<\:{x}_{{k}} \:<\:{x}_{{k}+\mathrm{1}} \\ $$$$.....,\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{1}}{\pi}\underset{{k}=\mathrm{1}} {\overset{\mathrm{1000}} {\sum}}\mid{x}_{{k}+\mathrm{1}} \:−\:{x}_{{k}} \mid\:\mathrm{is} \\ $$

Question Number 18891 Answers: 2 Comments: 0

If angles A and B satisfy (√2) cos A = cos B + cos^3 B and (√2) sin A = sin B − sin^3 B, then the value of 1620sin^2 (A − B) is

$$\mathrm{If}\:\mathrm{angles}\:{A}\:\mathrm{and}\:{B}\:\mathrm{satisfy}\:\sqrt{\mathrm{2}}\:\mathrm{cos}\:{A}\:= \\ $$$$\mathrm{cos}\:{B}\:+\:\mathrm{cos}^{\mathrm{3}} \:{B}\:\mathrm{and}\:\sqrt{\mathrm{2}}\:\mathrm{sin}\:{A}\:=\:\mathrm{sin}\:{B}\:− \\ $$$$\mathrm{sin}^{\mathrm{3}} \:{B},\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{1620sin}^{\mathrm{2}} \left({A}\:−\:{B}\right) \\ $$$$\mathrm{is} \\ $$

Question Number 18889 Answers: 0 Comments: 2

∫ ((d((√(3x))))/(√((√x) + 7)))

$$\int\:\frac{{d}\left(\sqrt{\mathrm{3}{x}}\right)}{\sqrt{\sqrt{{x}}\:+\:\mathrm{7}}} \\ $$

Question Number 18872 Answers: 2 Comments: 0

Question Number 18871 Answers: 0 Comments: 1

Question Number 18866 Answers: 1 Comments: 0

Question Number 18864 Answers: 2 Comments: 1

Question Number 18859 Answers: 1 Comments: 0

If f(x+2)=f(x)+7 and f(1)=2; f(2)=5, then the ratio of f(150) to f(75) is

$$\mathrm{If}\:\mathrm{f}\left(\mathrm{x}+\mathrm{2}\right)=\mathrm{f}\left(\mathrm{x}\right)+\mathrm{7}\:\mathrm{and}\:\mathrm{f}\left(\mathrm{1}\right)=\mathrm{2};\:\mathrm{f}\left(\mathrm{2}\right)=\mathrm{5}, \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{150}\right)\:\mathrm{to}\:\mathrm{f}\left(\mathrm{75}\right)\:\mathrm{is} \\ $$

Question Number 18854 Answers: 0 Comments: 0

lim_(x→0) sin x_x =?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}sin}\:\underset{\mathrm{x}} {\mathrm{x}}\:\:\:\:=? \\ $$

Question Number 18847 Answers: 1 Comments: 0

In an equilateral triangle with usual notations the value of ((27r^2 R)/(r_1 r_2 r_3 )) is equal to

$$\mathrm{In}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{usual} \\ $$$$\mathrm{notations}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{27}{r}^{\mathrm{2}} {R}}{{r}_{\mathrm{1}} {r}_{\mathrm{2}} {r}_{\mathrm{3}} }\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$

Question Number 18846 Answers: 0 Comments: 0

There are two crystalline forms of PbO ; one is yellow and the other is red. The standard enthalpies of formation of these two forms are −217.3 and −219 kJ per mole respectively. Calculate the enthalpy change for the solid-solid phase transition. PbO (yellow) → PbO (red)

$$\mathrm{There}\:\mathrm{are}\:\mathrm{two}\:\mathrm{crystalline}\:\mathrm{forms}\:\mathrm{of}\:\mathrm{PbO}\:; \\ $$$$\mathrm{one}\:\mathrm{is}\:\mathrm{yellow}\:\mathrm{and}\:\mathrm{the}\:\mathrm{other}\:\mathrm{is}\:\mathrm{red}.\:\mathrm{The} \\ $$$$\mathrm{standard}\:\mathrm{enthalpies}\:\mathrm{of}\:\mathrm{formation}\:\mathrm{of} \\ $$$$\mathrm{these}\:\mathrm{two}\:\mathrm{forms}\:\mathrm{are}\:−\mathrm{217}.\mathrm{3}\:\mathrm{and}\:−\mathrm{219} \\ $$$$\mathrm{kJ}\:\mathrm{per}\:\mathrm{mole}\:\mathrm{respectively}.\:\mathrm{Calculate}\:\mathrm{the} \\ $$$$\mathrm{enthalpy}\:\mathrm{change}\:\mathrm{for}\:\mathrm{the}\:\mathrm{solid}-\mathrm{solid} \\ $$$$\mathrm{phase}\:\mathrm{transition}. \\ $$$$\mathrm{PbO}\:\left(\mathrm{yellow}\right)\:\rightarrow\:\mathrm{PbO}\:\left(\mathrm{red}\right) \\ $$

Question Number 18885 Answers: 0 Comments: 0