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Question Number 103826 Answers: 1 Comments: 0

In the expansion of (1+x)^(20) if the coefficient of x^r is twice the coefficient of x^(r−1) , what the value of the coefficient?

$${In}\:{the}\:{expansion}\:{of}\:\left(\mathrm{1}+{x}\right)^{\mathrm{20}} \:{if}\:{the} \\ $$$${coefficient}\:{of}\:{x}^{{r}} \:{is}\:{twice}\:{the}\:{coefficient} \\ $$$${of}\:{x}^{{r}−\mathrm{1}} ,\:{what}\:{the}\:{value}\:{of}\:{the} \\ $$$${coefficient}?\: \\ $$

Question Number 103823 Answers: 2 Comments: 0

tan (x) = 4 cos (2x)−cot (2x)

$$\mathrm{tan}\:\left({x}\right)\:=\:\mathrm{4}\:\mathrm{cos}\:\left(\mathrm{2}{x}\right)−\mathrm{cot}\:\left(\mathrm{2}{x}\right) \\ $$

Question Number 103820 Answers: 1 Comments: 0

∫_R ^ (e^(−2iπax) /((1+x^2 )^2 ))dx = πe^(−2πa) ((1/2)+πa) a>0

$$\:\:\:\int_{\mathbb{R}} ^{} \:\frac{{e}^{−\mathrm{2}{i}\pi{ax}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:=\:\pi{e}^{−\mathrm{2}\pi{a}} \left(\frac{\mathrm{1}}{\mathrm{2}}+\pi{a}\right)\:\:\:\:\:\:\:\:\:{a}>\mathrm{0} \\ $$

Question Number 103887 Answers: 1 Comments: 3

how do you prove sin (a+b) = sin a cos b + cos a sin b geometrically ?

$${how}\:{do}\:{you}\:{prove}\:\mathrm{sin}\:\left({a}+{b}\right)\:=\:\mathrm{sin} \\ $$$${a}\:\mathrm{cos}\:{b}\:+\:\mathrm{cos}\:{a}\:\mathrm{sin}\:{b} \\ $$$${geometrically}\:? \\ $$

Question Number 103804 Answers: 0 Comments: 0

Solve for r ((h−p^2 )/(p−r))=(1/(2p)) , ((h−q)/(q^2 −r))= 2q (h−p^2 )^2 +(p−r)^2 =r^2 (h−q)^2 +(q^2 −r)^2 =r^2

$$\:\:\:\boldsymbol{{S}}{olve}\:{for}\:\boldsymbol{{r}} \\ $$$$\:\:\frac{\boldsymbol{{h}}−\boldsymbol{{p}}^{\mathrm{2}} }{\boldsymbol{{p}}−\boldsymbol{{r}}}=\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{{p}}}\:\:\:\:,\:\:\:\:\frac{\boldsymbol{{h}}−\boldsymbol{{q}}}{\boldsymbol{{q}}^{\mathrm{2}} −\boldsymbol{{r}}}=\:\mathrm{2}\boldsymbol{{q}} \\ $$$$\left(\boldsymbol{{h}}−\boldsymbol{{p}}^{\mathrm{2}} \right)^{\mathrm{2}} +\left(\boldsymbol{{p}}−\boldsymbol{{r}}\right)^{\mathrm{2}} =\boldsymbol{{r}}^{\mathrm{2}} \\ $$$$\:\left(\boldsymbol{{h}}−\boldsymbol{{q}}\right)^{\mathrm{2}} +\left(\boldsymbol{{q}}^{\mathrm{2}} −\boldsymbol{{r}}\right)^{\mathrm{2}} =\boldsymbol{{r}}^{\mathrm{2}} \\ $$

Question Number 103792 Answers: 2 Comments: 0

Question Number 103790 Answers: 1 Comments: 1

Question Number 103788 Answers: 1 Comments: 0

y′′−y = cot x

$${y}''−{y}\:=\:\mathrm{cot}\:{x}\: \\ $$

Question Number 105290 Answers: 2 Comments: 0

Given ((sin 2a−sin 2b)/(cos 2a+cos 2b)) = (2/3) find the value of cos (a−b)

$$\mathcal{G}{iven}\:\frac{\mathrm{sin}\:\mathrm{2}{a}−\mathrm{sin}\:\mathrm{2}{b}}{\mathrm{cos}\:\mathrm{2}{a}+\mathrm{cos}\:\mathrm{2}{b}}\:=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${find}\:{the}\:{value}\:{of}\:\mathrm{cos}\:\left({a}−{b}\right)\: \\ $$

Question Number 105289 Answers: 1 Comments: 0

(√(52))+(√3)−(√(4/6)) = ?

$$\sqrt{\mathrm{52}}+\sqrt{\mathrm{3}}−\sqrt{\frac{\mathrm{4}}{\mathrm{6}}}\:=\:? \\ $$

Question Number 103781 Answers: 0 Comments: 4

Question Number 103780 Answers: 1 Comments: 0

the third, sixth and seventh terms of a geometric progression (whose common ratio is neither 0 nor 1 ) are in arithmetic progression . prove that the sum of the first three terms is equal to fourth .

$${the}\:{third},\:{sixth}\:{and}\:{seventh}\:{terms}\:{of}\:{a} \\ $$$${geometric}\:{progression}\:\left({whose}\:{common}\right. \\ $$$$\left.{ratio}\:{is}\:{neither}\:\mathrm{0}\:{nor}\:\mathrm{1}\:\right)\:{are}\:{in} \\ $$$${arithmetic}\:{progression}\:.\:{prove}\:{that}\:{the} \\ $$$${sum}\:{of}\:{the}\:{first}\:{three}\:{terms}\:{is}\:{equal}\:{to} \\ $$$${fourth}\:. \\ $$

Question Number 103776 Answers: 2 Comments: 0

y′ − (y/(x^2 −1)) = y^2

$${y}'\:−\:\frac{{y}}{{x}^{\mathrm{2}} −\mathrm{1}}\:=\:{y}^{\mathrm{2}} \\ $$

Question Number 103774 Answers: 1 Comments: 0

Question Number 103773 Answers: 1 Comments: 0

∫_c ((x^2 +2xy^2 )dx+(x^2 y^2 −1)dy) where C is the boundary of region define by y^2 = 4x and y =1 ?

$$\int_{{c}} \left(\left({x}^{\mathrm{2}} +\mathrm{2}{xy}^{\mathrm{2}} \right){dx}+\left({x}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{1}\right){dy}\right) \\ $$$${where}\:{C}\:{is}\:{the}\:{boundary}\:{of} \\ $$$${region}\:{define}\:{by}\:{y}^{\mathrm{2}} =\:\mathrm{4}{x}\:{and}\:{y} \\ $$$$=\mathrm{1}\:? \\ $$

Question Number 103771 Answers: 0 Comments: 0

y′′−y′+y = cos 3x

$${y}''−{y}'+{y}\:=\:\mathrm{cos}\:\mathrm{3}{x} \\ $$

Question Number 103769 Answers: 1 Comments: 0

2y′′−y′+y = cos 3x

$$\mathrm{2}{y}''−{y}'+{y}\:=\:\mathrm{cos}\:\mathrm{3}{x} \\ $$

Question Number 103767 Answers: 2 Comments: 0

solve y′−y = y^4 at y(0) = 1

$${solve}\:{y}'−{y}\:=\:{y}^{\mathrm{4}} \:{at}\:{y}\left(\mathrm{0}\right)\:=\:\mathrm{1}\: \\ $$

Question Number 103766 Answers: 1 Comments: 0

given { ((x = ln 34)),((y = ln 38)) :} find ln 32 in terms of x and y

$${given}\:\begin{cases}{{x}\:=\:\mathrm{ln}\:\mathrm{34}}\\{{y}\:=\:\mathrm{ln}\:\mathrm{38}}\end{cases} \\ $$$${find}\:\mathrm{ln}\:\mathrm{32}\:{in}\:{terms}\:{of}\:{x}\:{and}\:{y}\: \\ $$

Question Number 103763 Answers: 1 Comments: 0

calculate{ Σ_(n=0) ^∞ (−1)^n x^n }×{Σ_(n=o) ^∞ (x^(2n) /(n+1))}

$$\mathrm{calculate}\left\{\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{x}^{\mathrm{n}} \right\}×\left\{\sum_{\mathrm{n}=\mathrm{o}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{2n}} }{\mathrm{n}+\mathrm{1}}\right\} \\ $$

Question Number 103759 Answers: 3 Comments: 0

(x^5 +3y) dx −x dy = 0

$$\left({x}^{\mathrm{5}} +\mathrm{3}{y}\right)\:{dx}\:−{x}\:{dy}\:=\:\mathrm{0}\: \\ $$

Question Number 103796 Answers: 1 Comments: 2

Σ_(k = 1) ^n k^5 = ?

$$\underset{{k}\:=\:\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{5}} \:=\:? \\ $$

Question Number 103756 Answers: 1 Comments: 1

Question Number 103755 Answers: 0 Comments: 0

Π_(n = 3) ^∞ (1−tan^4 ((π/2^n ))) = ?

$$\underset{{n}\:=\:\mathrm{3}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{4}} \left(\frac{\pi}{\mathrm{2}^{{n}} }\right)\right)\:=\:? \\ $$

Question Number 103753 Answers: 1 Comments: 0

→ { ((g(x)=(6/(x−2)))),(((goh)(3) = 17)),((h(x)= ax^2 −1)) :} find the value of a

$$\rightarrow\begin{cases}{{g}\left({x}\right)=\frac{\mathrm{6}}{{x}−\mathrm{2}}}\\{\left({goh}\right)\left(\mathrm{3}\right)\:=\:\mathrm{17}}\\{{h}\left({x}\right)=\:{ax}^{\mathrm{2}} −\mathrm{1}}\end{cases} \\ $$$${find}\:{the}\:{value}\:{of}\:{a}\: \\ $$

Question Number 103751 Answers: 1 Comments: 0

given that the parametric equation of a curvd are x=(1/(t−1)) y=(1/(t+1)) obtaun a cartesian equation of the curve. Hemce find an equqtion of tbe nirmal to the curve at the point t=2

$$ \\ $$$${given}\:{that}\:{the}\:{parametric}\:{equation}\:{of}\: \\ $$$${a}\:{curvd}\:{are}\:{x}=\frac{\mathrm{1}}{{t}−\mathrm{1}}\:\:{y}=\frac{\mathrm{1}}{{t}+\mathrm{1}}\:{obtaun}\:{a}\: \\ $$$${cartesian}\:{equation}\:{of}\:{the}\:{curve}.\:{Hemce}\:{find} \\ $$$${an}\:{equqtion}\:{of}\:{tbe}\:{nirmal}\:{to}\:{the}\:{curve}\:{at}\:{the}\: \\ $$$${point}\:{t}=\mathrm{2} \\ $$

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