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Question Number 157455 Answers: 1 Comments: 0

Question Number 157454 Answers: 2 Comments: 0

Given 1+(3/y)+(5/y^2 )+(7/y^3 )+(9/y^4 )+...= 71 then 1+(4/y)+(9/y^2 )+((16)/y^3 )+((25)/y^4 )+... =?

$$\:{Given}\:\mathrm{1}+\frac{\mathrm{3}}{{y}}+\frac{\mathrm{5}}{{y}^{\mathrm{2}} }+\frac{\mathrm{7}}{{y}^{\mathrm{3}} }+\frac{\mathrm{9}}{{y}^{\mathrm{4}} }+...=\:\mathrm{71} \\ $$$$\:{then}\:\mathrm{1}+\frac{\mathrm{4}}{{y}}+\frac{\mathrm{9}}{{y}^{\mathrm{2}} }+\frac{\mathrm{16}}{{y}^{\mathrm{3}} }+\frac{\mathrm{25}}{{y}^{\mathrm{4}} }+...\:=? \\ $$

Question Number 157453 Answers: 1 Comments: 0

Question Number 157452 Answers: 0 Comments: 1

solve tan x=cos 3x

$$\:{solve}\:\mathrm{tan}\:{x}=\mathrm{cos}\:\mathrm{3}{x}\: \\ $$

Question Number 157442 Answers: 2 Comments: 0

∫ (dx/(x−(√(x^2 +2x+2))))

$$\:\:\int\:\frac{\mathrm{dx}}{\mathrm{x}−\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{2}}}\: \\ $$

Question Number 157441 Answers: 2 Comments: 0

(7 ((6x−10))^(1/5) )^2 = ((49)/( ((1/x^2 ))^(1/5) ))

$$\:\left(\mathrm{7}\:\sqrt[{\mathrm{5}}]{\mathrm{6x}−\mathrm{10}}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{49}}{\:\sqrt[{\mathrm{5}}]{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}}\: \\ $$

Question Number 157425 Answers: 1 Comments: 0

Question Number 157420 Answers: 2 Comments: 0

∫(1/(u^4 +(1−u)^4 ))du

$$\int\frac{\mathrm{1}}{\boldsymbol{{u}}^{\mathrm{4}} +\left(\mathrm{1}−\boldsymbol{{u}}\right)^{\mathrm{4}} }\boldsymbol{{du}} \\ $$

Question Number 157418 Answers: 0 Comments: 0

Question Number 157417 Answers: 1 Comments: 0

if a;b>0 then prove that: (((a + b)^6 (a^(15) + b^(15) ) (a^(21) + b^(21) ))/((a^3 + b^3 )^2 (a^5 + b^5 )^3 (a^7 + b^7 )^3 )) ≥ 1

$$\mathrm{if}\:\:\mathrm{a};\mathrm{b}>\mathrm{0}\:\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\left(\mathrm{a}\:+\:\mathrm{b}\right)^{\mathrm{6}} \:\left(\mathrm{a}^{\mathrm{15}} \:+\:\mathrm{b}^{\mathrm{15}} \right)\:\left(\mathrm{a}^{\mathrm{21}} \:+\:\mathrm{b}^{\mathrm{21}} \right)}{\left(\mathrm{a}^{\mathrm{3}} \:+\:\mathrm{b}^{\mathrm{3}} \right)^{\mathrm{2}} \:\left(\mathrm{a}^{\mathrm{5}} \:+\:\mathrm{b}^{\mathrm{5}} \right)^{\mathrm{3}} \:\left(\mathrm{a}^{\mathrm{7}} \:+\:\mathrm{b}^{\mathrm{7}} \right)^{\mathrm{3}} }\:\geqslant\:\mathrm{1} \\ $$$$ \\ $$

Question Number 157416 Answers: 3 Comments: 3

If 2x^2 +11x+5 is a factor of ax^3 −17x^2 +bx−15 , then what is a and b ?

$$\:{If}\:\mathrm{2}{x}^{\mathrm{2}} +\mathrm{11}{x}+\mathrm{5}\:{is}\:{a}\:{factor}\:{of} \\ $$$$\:{ax}^{\mathrm{3}} −\mathrm{17}{x}^{\mathrm{2}} +{bx}−\mathrm{15}\:,\:{then}\:{what} \\ $$$${is}\:{a}\:{and}\:{b}\:? \\ $$

Question Number 157412 Answers: 2 Comments: 0

∫_( 0) ^( (π/2)) (x/(sin^8 x+cos^8 x)) dx ?

$$\:\int_{\:\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\frac{{x}}{\mathrm{sin}\:^{\mathrm{8}} {x}+\mathrm{cos}\:^{\mathrm{8}} {x}}\:{dx}\:? \\ $$

Question Number 157392 Answers: 1 Comments: 2

Question Number 157389 Answers: 0 Comments: 0

lim_(x→∞) (((6e^x^2 −20ln ∣x∣)/(6e^x^2 −20ln ∣x∣+3)))^(3e^(2x^2 ) sin (((6k)/e^x^2 ))) =26 k=?

$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{6}{e}^{{x}^{\mathrm{2}} } −\mathrm{20ln}\:\mid{x}\mid}{\mathrm{6}{e}^{{x}^{\mathrm{2}} } −\mathrm{20ln}\:\mid{x}\mid+\mathrm{3}}\right)^{\mathrm{3}{e}^{\mathrm{2}{x}^{\mathrm{2}} } \:\mathrm{sin}\:\left(\frac{\mathrm{6}{k}}{{e}^{{x}^{\mathrm{2}} } }\right)} =\mathrm{26} \\ $$$$\:{k}=? \\ $$$$ \\ $$

Question Number 157388 Answers: 1 Comments: 2

If m tan (θ − 30°) = n tan (θ + 12°) cos 2θ = ?

$${If}\:\:{m}\:\mathrm{tan}\:\left(\theta\:−\:\mathrm{30}°\right)\:=\:{n}\:\mathrm{tan}\:\left(\theta\:+\:\mathrm{12}°\right) \\ $$$$\mathrm{cos}\:\mathrm{2}\theta\:\:=\:? \\ $$

Question Number 157382 Answers: 1 Comments: 0

Question Number 157379 Answers: 2 Comments: 1

Question Number 157374 Answers: 0 Comments: 3

I see there are idiot comments on this forum. a comment that tells someone not to answer a member's question. this is a provocateur. what's the point of telling people not to answer questions posted by members. If the one who ordered it doesn't want to answer, then do it. no need to tell people to follow you.

$$ \\ $$I see there are idiot comments on this forum. a comment that tells someone not to answer a member's question. this is a provocateur. what's the point of telling people not to answer questions posted by members. If the one who ordered it doesn't want to answer, then do it. no need to tell people to follow you.

Question Number 157351 Answers: 3 Comments: 1

Prove by mathematical induction Σ_(r=1) ^n (1/(r(r+1))) = (n/(n+1))

$$\mathrm{Prove}\:\mathrm{by}\:\mathrm{mathematical}\:\mathrm{induction} \\ $$$$\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{r}\left({r}+\mathrm{1}\right)}\:=\:\frac{{n}}{{n}+\mathrm{1}} \\ $$

Question Number 157343 Answers: 1 Comments: 0