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AllQuestion and Answers: Page 1080

Question Number 103093 Answers: 5 Comments: 0

solve for x,y∈N (1/x)+(1/y)=(3/(202))

$${solve}\:{for}\:{x},{y}\in\mathbb{N} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}=\frac{\mathrm{3}}{\mathrm{202}} \\ $$

Question Number 102980 Answers: 1 Comments: 6

Question Number 102978 Answers: 2 Comments: 1

Question Number 102975 Answers: 0 Comments: 1

A student can recall 6 digits of a 9 digit number. In how many ways can he get the complete number?

$$\mathrm{A}\:\mathrm{student}\:\mathrm{can}\:\mathrm{recall}\:\mathrm{6}\:\mathrm{digits}\:\mathrm{of}\:\mathrm{a}\:\mathrm{9}\:\mathrm{digit}\:\mathrm{number}. \\ $$$$\mathrm{In}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{he}\:\mathrm{get}\:\mathrm{the}\:\mathrm{complete}\:\mathrm{number}? \\ $$

Question Number 102967 Answers: 1 Comments: 0

Question Number 102944 Answers: 0 Comments: 5

Question Number 102940 Answers: 0 Comments: 1

Question Number 102927 Answers: 3 Comments: 1

If (x/(x^2 + x + 1)) = (1/4) Find x + (1/x)

$$\mathrm{If}\:\:\:\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{2}} \:\:+\:\:\mathrm{x}\:\:+\:\:\mathrm{1}}\:\:\:=\:\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{Find}\:\:\:\:\:\mathrm{x}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{x}} \\ $$

Question Number 102926 Answers: 3 Comments: 0

∫ (dθ/(2sin^2 θ−cos^2 θ)) ?

$$\:\int\:\frac{{d}\theta}{\mathrm{2}{sin}^{\mathrm{2}} \theta−{cos}^{\mathrm{2}} \theta}\:\:? \\ $$

Question Number 102922 Answers: 1 Comments: 1

∫ (dx/(x^3 +3x−5)) ?

$$\int\:\frac{{dx}}{{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{5}}\:\:? \\ $$

Question Number 103071 Answers: 0 Comments: 0

∫(e^x /((1+x^2 )^2 ))∙(x^3 −x^2 +x+1)dx

$$\int\frac{\mathrm{e}^{\mathrm{x}} }{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\centerdot\left(\mathrm{x}^{\mathrm{3}} −\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)\mathrm{dx} \\ $$

Question Number 102918 Answers: 1 Comments: 2

Question Number 102911 Answers: 2 Comments: 1

∫ ((sin(x))/x) dx

$$\int\:\frac{\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{x}}\:\:\mathrm{dx} \\ $$

Question Number 102910 Answers: 1 Comments: 0

(1)Σ_(m = 1) ^n tan^2 (((mπ)/(2n+1))) (2) Π_(m = 1) ^n tan^2 (((mπ)/(2n+1)))

$$\left(\mathrm{1}\right)\underset{\mathrm{m}\:=\:\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{m}\pi}{\mathrm{2n}+\mathrm{1}}\right) \\ $$$$\left(\mathrm{2}\right)\:\underset{\mathrm{m}\:=\:\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{m}\pi}{\mathrm{2n}+\mathrm{1}}\right)\: \\ $$

Question Number 102905 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 102880 Answers: 1 Comments: 5

Question Number 102963 Answers: 0 Comments: 0

Question Number 102857 Answers: 0 Comments: 1

Question Number 102854 Answers: 1 Comments: 0

Question Number 102883 Answers: 1 Comments: 0

If x^3 +ax^2 +bx+c = 0 has the roots are α^ β and γ . find the value of αβ^2 +βγ^2 +γα^2 in terms a,b and c

$${If}\:{x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c}\:=\:\mathrm{0}\:{has}\:{the}\:{roots}\:{are}\: \\ $$$$\bar {\alpha}\:\beta\:{and}\:\gamma\:.\:{find}\:{the}\:{value}\:{of} \\ $$$$\alpha\beta^{\mathrm{2}} +\beta\gamma^{\mathrm{2}} +\gamma\alpha^{\mathrm{2}} \:{in}\:{terms}\:{a},{b}\:{and}\:{c} \\ $$

Question Number 102843 Answers: 0 Comments: 8

Question Number 102840 Answers: 1 Comments: 0

y′+(√(x+y−1))=x+y+1

$${y}'+\sqrt{{x}+{y}−\mathrm{1}}={x}+{y}+\mathrm{1} \\ $$

Question Number 102838 Answers: 1 Comments: 2

Question Number 102836 Answers: 2 Comments: 1

Question Number 102822 Answers: 1 Comments: 1

Σ_(n=1) ^∞ ((n!)/n^n )

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}!}{{n}^{{n}} } \\ $$

Question Number 102819 Answers: 0 Comments: 4

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