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Question Number 102796 Answers: 1 Comments: 0

The angle between the vectors 2i+3j+k and 2i−j−k is

$$\mathrm{The}\:\mathrm{angle}\:\mathrm{between}\:\mathrm{the}\:\mathrm{vectors} \\ $$$$\mathrm{2}\boldsymbol{\mathrm{i}}+\mathrm{3}\boldsymbol{\mathrm{j}}+\boldsymbol{\mathrm{k}}\:\:\mathrm{and}\:\:\mathrm{2}\boldsymbol{\mathrm{i}}−\boldsymbol{\mathrm{j}}−\boldsymbol{\mathrm{k}}\:\:\mathrm{is} \\ $$

Question Number 102795 Answers: 0 Comments: 0

Points D, E are taken on the side BC of a △ ABC, such that BD=DE=EC. If ∠BAD=x, ∠DAE=y, ∠EAC=z, then the value of ((sin (x+y) sin(y+z))/(sin x sin z)) is equal to

$$\mathrm{Points}\:{D},\:{E}\:\mathrm{are}\:\mathrm{taken}\:\mathrm{on}\:\mathrm{the}\:\mathrm{side}\:{BC} \\ $$$$\mathrm{of}\:\mathrm{a}\:\bigtriangleup\:{ABC},\:\mathrm{such}\:\mathrm{that}\:{BD}={DE}={EC}. \\ $$$$\mathrm{If}\:\:\angle{BAD}={x},\:\angle{DAE}={y},\:\angle{EAC}={z}, \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\:\frac{\mathrm{sin}\:\left({x}+{y}\right)\:\mathrm{sin}\left({y}+{z}\right)}{\mathrm{sin}\:{x}\:\mathrm{sin}\:{z}} \\ $$$$\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$

Question Number 102790 Answers: 1 Comments: 0

I=2∫_0 ^(1/(√2)) ((sin^(−1) (x))/x) dx −∫_0 ^1 ((tan^(−1) (x))/x) dx

$${I}=\mathrm{2}\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} {\int}}\:\frac{\mathrm{sin}^{−\mathrm{1}} \left({x}\right)}{{x}}\:{dx}\:−\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}{{x}}\:{dx}\: \\ $$

Question Number 102784 Answers: 3 Comments: 0

Σ_(n=1) ^∞ (n/4^n ) =?

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{n}}{\mathrm{4}^{{n}} }\:=? \\ $$

Question Number 102783 Answers: 2 Comments: 0

2(√3)−1 = 6sin (2θ−60^o )−2sin (2θ−30^o ) ⇒θ=?

$$\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}\:=\:\mathrm{6sin}\:\left(\mathrm{2}\theta−\mathrm{60}^{\mathrm{o}} \right)−\mathrm{2sin} \\ $$$$\left(\mathrm{2}\theta−\mathrm{30}^{\mathrm{o}} \right)\:\Rightarrow\theta=? \\ $$

Question Number 102773 Answers: 0 Comments: 0

During a sales period a magazine offers a t% discount For clients in possession of the fidelity card, an extra discount of (t+5)% is offered. A client benefits from these two discounts and pays 150 ε for an article whose initial price is 250ε (i) Show that t is solution to the equation 250×(1−(t/(100)))×(1−((t+5)/(100)))=150 (ii) Solve this equation and deduce the value of t.

$$\:\:\:\mathrm{During}\:\mathrm{a}\:\mathrm{sales}\:\mathrm{period}\:\mathrm{a}\:\mathrm{magazine}\:\mathrm{offers}\:\mathrm{a}\:\mathrm{t\%}\:\mathrm{discount} \\ $$$$\mathrm{For}\:\mathrm{clients}\:\mathrm{in}\:\mathrm{possession}\:\mathrm{of}\:\mathrm{the}\:\mathrm{fidelity}\:\mathrm{card},\:\mathrm{an}\:\mathrm{extra}\:\mathrm{discount} \\ $$$$\mathrm{of}\:\left(\mathrm{t}+\mathrm{5}\right)\%\:\mathrm{is}\:\mathrm{offered}. \\ $$$$\:\:\:\mathrm{A}\:\mathrm{client}\:\mathrm{benefits}\:\mathrm{from}\:\mathrm{these}\:\mathrm{two}\:\mathrm{discounts}\:\mathrm{and}\:\mathrm{pays}\: \\ $$$$\mathrm{150}\:\epsilon\:\mathrm{for}\:\mathrm{an}\:\mathrm{article}\:\mathrm{whose}\:\mathrm{initial}\:\mathrm{price}\:\mathrm{is}\:\mathrm{250}\epsilon \\ $$$$\left({i}\right)\:\mathrm{Show}\:\mathrm{that}\:\mathrm{t}\:\mathrm{is}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{250}×\left(\mathrm{1}−\frac{\mathrm{t}}{\mathrm{100}}\right)×\left(\mathrm{1}−\frac{\mathrm{t}+\mathrm{5}}{\mathrm{100}}\right)=\mathrm{150} \\ $$$$\left({ii}\right)\:\mathrm{Solve}\:\mathrm{this}\:\mathrm{equation}\:\mathrm{and}\:\mathrm{deduce}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{t}. \\ $$

Question Number 102769 Answers: 2 Comments: 9

Question Number 102771 Answers: 2 Comments: 0

x+(1/x) = −1 ⇒x^(1907) +(1/x^(1907) ) ?

$$\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\:=\:−\mathrm{1}\:\Rightarrow\mathrm{x}^{\mathrm{1907}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{1907}} }\:?\: \\ $$

Question Number 102894 Answers: 3 Comments: 0

∫ (√(x+(√(x+(√(x+(√(x+(√(x+...)))))))))) dx

$$\int\:\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+...}}}}}\:\mathrm{dx}\: \\ $$

Question Number 102763 Answers: 0 Comments: 0

Show that the function defined within [0,1] by f(x)= { ((1 if x∈Q∩[0,1])),((0 otherwise)) :} is not Riemann integrable within [0,1]

$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{function}\:\mathrm{defined}\:\mathrm{within}\:\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\mathrm{by}\:\mathrm{f}\left(\mathrm{x}\right)=\begin{cases}{\mathrm{1}\:\mathrm{if}\:\mathrm{x}\in\mathbb{Q}\cap\left[\mathrm{0},\mathrm{1}\right]}\\{\mathrm{0}\:\mathrm{otherwise}}\end{cases}\:\:\mathrm{is}\:\mathrm{not}\:\mathrm{Riemann}\:\mathrm{integrable} \\ $$$$\mathrm{within}\:\left[\mathrm{0},\mathrm{1}\right] \\ $$

Question Number 102753 Answers: 1 Comments: 0

A wedge has a weight of 9kg . And a block has a weight of 2kg If the block starts sliding with an angle of 45° with the horizontal then what is accelaration of the wedge?

$${A}\:{wedge}\:{has}\:{a}\:{weight}\:{of}\:\mathrm{9}{kg}\:.\:{And}\:{a}\:{block}\:{has}\:{a}\:{weight}\:{of}\:\mathrm{2}{kg} \\ $$$${If}\:{the}\:{block}\:{starts}\:{sliding}\:{with}\:{an}\:{angle}\:{of}\:\mathrm{45}°\:{with}\:{the} \\ $$$${horizontal}\:{then}\:{what}\:{is}\:{accelaration}\:{of}\:{the}\:{wedge}? \\ $$$$ \\ $$$$ \\ $$

Question Number 102745 Answers: 0 Comments: 2

1−1+1−1+1−1+1−1+.....=(1/2) {But it diverges 1+1+1+1+1+1+1+......=−(1/2) {But it diverges 1+2+4+8+16+.....=−1 {But it diverges 1+2+3+4+5+6+7=−(1/(12)) {But it diverges 1−2+4−8+.....=(1/3) {But it diverges 1−2+3−4+5−6+.....=(1/4) {Is it a divergent?????

$$\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+.....=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\left\{{But}\:{it}\:{diverges}\right. \\ $$$$\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+......=−\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\left\{{But}\:{it}\:{diverges}\right. \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{4}+\mathrm{8}+\mathrm{16}+.....=−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{{But}\:{it}\:{diverges}\right. \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+\mathrm{7}=−\frac{\mathrm{1}}{\mathrm{12}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{{But}\:{it}\:{diverges}\right. \\ $$$$\mathrm{1}−\mathrm{2}+\mathrm{4}−\mathrm{8}+.....=\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{{But}\:{it}\:{diverges}\right. \\ $$$$\mathrm{1}−\mathrm{2}+\mathrm{3}−\mathrm{4}+\mathrm{5}−\mathrm{6}+.....=\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\:\:\left\{{Is}\:{it}\:{a}\:{divergent}?????\right. \\ $$

Question Number 102721 Answers: 1 Comments: 1

f(((100x−1)/(50x+1))) = 2x−1 & f^(−1) (3)= p p=?

$$\mathrm{f}\left(\frac{\mathrm{100x}−\mathrm{1}}{\mathrm{50x}+\mathrm{1}}\right)\:=\:\mathrm{2x}−\mathrm{1}\:\&\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{3}\right)=\:\mathrm{p} \\ $$$$\mathrm{p}=? \\ $$

Question Number 102716 Answers: 3 Comments: 0

y′′+3y′−10y=14e^(−5x)

$$\mathrm{y}''+\mathrm{3y}'−\mathrm{10y}=\mathrm{14e}^{−\mathrm{5x}} \\ $$

Question Number 102714 Answers: 0 Comments: 1