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Question Number 168642 Answers: 0 Comments: 0

Let △ABC be a triangle with ∠ABC=60° and ∠ACB=50°. IABI=a^2 −2 , IBCI=a in this instance , prove that IACI=(√3).

$$\mathrm{Let}\:\:\bigtriangleup\mathrm{ABC}\:\:\mathrm{be}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with} \\ $$$$\angle\mathrm{ABC}=\mathrm{60}°\:\:\mathrm{and}\:\:\angle\mathrm{ACB}=\mathrm{50}°. \\ $$$$\mathrm{IABI}=\mathrm{a}^{\mathrm{2}} −\mathrm{2}\:,\:\mathrm{IBCI}=\mathrm{a}\:\:\mathrm{in}\:\mathrm{this} \\ $$$$\mathrm{instance}\:,\:\mathrm{prove}\:\mathrm{that}\:\:\mathrm{IACI}=\sqrt{\mathrm{3}}. \\ $$

Question Number 168628 Answers: 1 Comments: 0

montrer que d(x,y)=((∣u−v∣)/(1+∣u−v∣)) une distance sur R

$${montrer}\:{que} \\ $$$${d}\left({x},{y}\right)=\frac{\mid{u}−{v}\mid}{\mathrm{1}+\mid{u}−{v}\mid}\: \\ $$$${une}\:{distance}\:{sur}\:{R} \\ $$

Question Number 168617 Answers: 0 Comments: 2

Question Number 168616 Answers: 1 Comments: 0

Question Number 168613 Answers: 3 Comments: 1

Resolve 1) x(dy/dx)−y=y^3 2) (x−y)ydx−x^2 dy=0 3) (2x−y)dx+(4x−2y+3)dy=0

$${Resolve}\: \\ $$$$\left.\mathrm{1}\right)\:{x}\frac{{dy}}{{dx}}−{y}={y}^{\mathrm{3}} \\ $$$$\left.\mathrm{2}\right)\:\left({x}−{y}\right){ydx}−{x}^{\mathrm{2}} {dy}=\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\:\left(\mathrm{2}{x}−{y}\right){dx}+\left(\mathrm{4}{x}−\mathrm{2}{y}+\mathrm{3}\right){dy}=\mathrm{0} \\ $$

Question Number 168609 Answers: 2 Comments: 1

Question Number 168608 Answers: 2 Comments: 0

Question Number 168606 Answers: 1 Comments: 6

Question Number 168605 Answers: 0 Comments: 4

((cos^2 10°+sin^2 25°−cos^2 15°)/(sin^2 10°+sin^2 25°−sin^2 15°))=?

$$\:\:\:\:\:\frac{\mathrm{cos}\:^{\mathrm{2}} \mathrm{10}°+\mathrm{sin}\:^{\mathrm{2}} \mathrm{25}°−\mathrm{cos}\:^{\mathrm{2}} \mathrm{15}°}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{10}°+\mathrm{sin}\:^{\mathrm{2}} \mathrm{25}°−\mathrm{sin}\:^{\mathrm{2}} \mathrm{15}°}=? \\ $$

Question Number 168603 Answers: 1 Comments: 1

Question Number 168595 Answers: 0 Comments: 1

Question Number 168593 Answers: 0 Comments: 3

Question Number 168585 Answers: 1 Comments: 2

if: U_0 =0 , U_(n+1) = (2n+2)U_n +2n+1 find U_(n ) ?

$${if}:\:{U}_{\mathrm{0}} =\mathrm{0}\:,\:{U}_{{n}+\mathrm{1}} \:=\:\left(\mathrm{2}{n}+\mathrm{2}\right){U}_{{n}} +\mathrm{2}{n}+\mathrm{1}\:{find}\:{U}_{{n}\:} \:? \\ $$

Question Number 168579 Answers: 0 Comments: 1

Question Number 168576 Answers: 0 Comments: 1

If lim_(x→0) ((cos^m (mx)−cos^n (nx))/((m^2 +n^2 +mn)x^2 )) = 1 find ((m^2 +n^2 −4)/(mn)) .

$$\:\:\:{If}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:^{{m}} \left({mx}\right)−\mathrm{cos}\:^{{n}} \left({nx}\right)}{\left({m}^{\mathrm{2}} +{n}^{\mathrm{2}} +{mn}\right){x}^{\mathrm{2}} \:}\:=\:\mathrm{1} \\ $$$$\:{find}\:\frac{{m}^{\mathrm{2}} +{n}^{\mathrm{2}} −\mathrm{4}}{{mn}}\:. \\ $$

Question Number 168575 Answers: 1 Comments: 0

Calculate :: lim_(x→+∞) (((x+a)^(x+a) (x+b)^(x+b) )/((x+a+b)^(2x+a+b) ))=?

$$\mathrm{Calculate}\:::\:\:\underset{\mathrm{x}\rightarrow+\infty} {\mathrm{lim}}\frac{\left(\mathrm{x}+\mathrm{a}\right)^{\mathrm{x}+\mathrm{a}} \left(\mathrm{x}+\mathrm{b}\right)^{\mathrm{x}+\mathrm{b}} }{\left(\mathrm{x}+\mathrm{a}+\mathrm{b}\right)^{\mathrm{2x}+\mathrm{a}+\mathrm{b}} }=? \\ $$

Question Number 168558 Answers: 2 Comments: 1

Question Number 168556 Answers: 1 Comments: 0

∫(√(sinx)) dx

$$\int\sqrt{{sinx}}\:{dx} \\ $$

Question Number 168555 Answers: 2 Comments: 0

Question Number 168552 Answers: 0 Comments: 0

Question Number 168550 Answers: 1 Comments: 0

Question Number 168549 Answers: 1 Comments: 0

Resolve (x−2)^2 y^(′′) −3(x−2)y′+y=x

$${Resolve} \\ $$$$\left({x}−\mathrm{2}\right)^{\mathrm{2}} {y}^{''} −\mathrm{3}\left({x}−\mathrm{2}\right){y}'+{y}={x} \\ $$

Question Number 168548 Answers: 1 Comments: 0

Question Number 168546 Answers: 0 Comments: 0

Question Number 168538 Answers: 2 Comments: 0

Question Number 168537 Answers: 2 Comments: 0

4^(61) +4^(62) +4^(63) +4^(64 ) is divisible by (1) 17 (2) 3 (3) 11 (4) 13 Mastermind

$$\mathrm{4}^{\mathrm{61}} +\mathrm{4}^{\mathrm{62}} +\mathrm{4}^{\mathrm{63}} +\mathrm{4}^{\mathrm{64}\:} \:{is}\:{divisible}\:{by} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{17}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\right)\:\mathrm{3} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{11}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{4}\right)\:\mathrm{13} \\ $$$$ \\ $$$${Mastermind} \\ $$

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