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AllQuestion and Answers: Page 9

Question Number 225356 Answers: 4 Comments: 1

Question Number 225354 Answers: 1 Comments: 1

Question Number 225338 Answers: 1 Comments: 1

Question Number 225330 Answers: 1 Comments: 0

if (fogoh)(x)=cos^2 (x+9) then f(x)=? , g(x)=? , h(x)=?

$${if}\:\:\left({fogoh}\right)\left({x}\right)={cos}^{\mathrm{2}} \left({x}+\mathrm{9}\right) \\ $$$${then}\:\:\:{f}\left({x}\right)=?\:,\:\:{g}\left({x}\right)=?\:\:,\:{h}\left({x}\right)=? \\ $$

Question Number 225323 Answers: 1 Comments: 0

Find: (((3 + 2 (5)^(1/4) )/(3 - 2 (5)^(1/4) )))^(1/4) . (((5)^(1/4) - 1)/( (5)^(1/4) + 1)) = ?

$$\mathrm{Find}:\:\:\:\left(\frac{\mathrm{3}\:+\:\mathrm{2}\:\sqrt[{\mathrm{4}}]{\mathrm{5}}}{\mathrm{3}\:-\:\mathrm{2}\:\sqrt[{\mathrm{4}}]{\mathrm{5}}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} .\:\:\:\frac{\sqrt[{\mathrm{4}}]{\mathrm{5}}\:-\:\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{5}}\:+\:\mathrm{1}}\:\:=\:? \\ $$

Question Number 225394 Answers: 1 Comments: 2

Question Number 225307 Answers: 1 Comments: 1

Question Number 225299 Answers: 0 Comments: 1

We have integrated artificial intelligence: - images are checked during upload and will be rejected if not related to science - daily processing of other posts A few senior users are exempted from these checks.

$$\mathrm{We}\:\mathrm{have}\:\mathrm{integrated}\:\mathrm{artificial} \\ $$$$\mathrm{intelligence}: \\ $$$$-\:\mathrm{images}\:\mathrm{are}\:\mathrm{checked}\:\mathrm{during}\:\mathrm{upload} \\ $$$$\:\:\:\mathrm{and}\:\mathrm{will}\:\mathrm{be}\:\mathrm{rejected}\:\mathrm{if}\:\mathrm{not}\:\mathrm{related} \\ $$$$\:\:\:\mathrm{to}\:\mathrm{science} \\ $$$$-\:\mathrm{daily}\:\mathrm{processing}\:\mathrm{of}\:\mathrm{other}\:\mathrm{posts} \\ $$$$\mathrm{A}\:\mathrm{few}\:\mathrm{senior}\:\mathrm{users}\:\mathrm{are}\:\mathrm{exempted}\:\mathrm{from} \\ $$$$\mathrm{these}\:\mathrm{checks}. \\ $$

Question Number 225303 Answers: 0 Comments: 0

Let S_n (x)=Σ_(r=1) ^n ((sin ((2r−3)2^(−(r+1)) x)cos ((10r+1)2^(−(r+1)) x)−sin( (6r−1)2^(−(r+1)) x)cos ((2r+5)2^(−(r+1)) x))/(2^(r−1) (sin (r2^(3−r) −x)sin (2^(2−r) −x)))) then find the value of lim_(m→∞) (((Σ_(n=0) ^m ∫_0 ^1 cos^(−1) (((cos (x))/(2^n (1+2S_n (x)cos (x)))))dx)/([(d/dx)(Σ_(n=0) ^m cos^(−1) (((cos (x))/(2^n (1+2S_n (x)cos (x)))))]_(x=0) )))

$${Let} \\ $$$${S}_{{n}} \left({x}\right)=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{sin}\:\left(\left(\mathrm{2}{r}−\mathrm{3}\right)\mathrm{2}^{−\left({r}+\mathrm{1}\right)} {x}\right)\mathrm{cos}\:\left(\left(\mathrm{10}{r}+\mathrm{1}\right)\mathrm{2}^{−\left({r}+\mathrm{1}\right)} {x}\right)−\mathrm{sin}\left(\:\left(\mathrm{6}{r}−\mathrm{1}\right)\mathrm{2}^{−\left({r}+\mathrm{1}\right)} {x}\right)\mathrm{cos}\:\left(\left(\mathrm{2}{r}+\mathrm{5}\right)\mathrm{2}^{−\left({r}+\mathrm{1}\right)} {x}\right)}{\mathrm{2}^{{r}−\mathrm{1}} \left(\mathrm{sin}\:\left({r}\mathrm{2}^{\mathrm{3}−{r}} −{x}\right)\mathrm{sin}\:\left(\mathrm{2}^{\mathrm{2}−{r}} −{x}\right)\right)} \\ $$$${then}\:{find}\:{the}\:{value}\:{of} \\ $$$$\underset{{m}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\underset{{n}=\mathrm{0}} {\overset{{m}} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{cos}\:\left({x}\right)}{\mathrm{2}^{{n}} \left(\mathrm{1}+\mathrm{2}{S}_{{n}} \left({x}\right)\mathrm{cos}\:\left({x}\right)\right)}\right){dx}}{\left[\frac{{d}}{{dx}}\left(\underset{{n}=\mathrm{0}} {\overset{{m}} {\sum}}\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{cos}\:\left({x}\right)}{\mathrm{2}^{{n}} \left(\mathrm{1}+\mathrm{2}{S}_{{n}} \left({x}\right)\mathrm{cos}\:\left({x}\right)\right)}\right)\right]_{{x}=\mathrm{0}} \right.}\right) \\ $$

Question Number 225282 Answers: 2 Comments: 4

Q225146 here if the wedge and the ground had no friction it would start going → what is acc. at time t? μ=kx

$${Q}\mathrm{225146} \\ $$$${here}\:{if}\:{the}\:{wedge}\:{and}\:{the} \\ $$$${ground}\:\:{had}\:{no}\:{friction} \\ $$$${it}\:{would}\:{start}\:{going}\:\rightarrow \\ $$$${what}\:{is}\:{acc}.\:{at}\:{time}\:{t}? \\ $$$$\mu={kx} \\ $$

Question Number 225230 Answers: 1 Comments: 3

tg^4 10°+tg^4 50°+tg^4 70°=?

$$\:\:\:{tg}^{\mathrm{4}} \mathrm{10}°+{tg}^{\mathrm{4}} \mathrm{50}°+{tg}^{\mathrm{4}} \mathrm{70}°=? \\ $$

Question Number 225240 Answers: 1 Comments: 0

Question Number 225239 Answers: 0 Comments: 0

Question Number 225241 Answers: 1 Comments: 0

Question Number 225153 Answers: 0 Comments: 0

Question Number 225152 Answers: 0 Comments: 0

Question Number 225151 Answers: 0 Comments: 0

Question Number 225150 Answers: 0 Comments: 0

Question Number 225146 Answers: 2 Comments: 9

Question Number 225075 Answers: 2 Comments: 0

Question Number 225072 Answers: 1 Comments: 0

Question Number 225098 Answers: 2 Comments: 0

Question Number 225054 Answers: 3 Comments: 2

Question Number 225047 Answers: 2 Comments: 0

∫ x^x dx

$$\int\:{x}^{{x}} \:{dx} \\ $$

Question Number 225020 Answers: 0 Comments: 1

tgx+tgy+tgz=A tg^3 x+tg^3 y+tg^3 z=?

$$\:\:\:{tgx}+{tgy}+{tgz}={A} \\ $$$$\:\:\:{tg}^{\mathrm{3}} {x}+{tg}^{\mathrm{3}} {y}+{tg}^{\mathrm{3}} {z}=? \\ $$

Question Number 225003 Answers: 1 Comments: 17

The new symbols are in progress. New update will be release by end of this month with - Close integration symbols - Italic greek capital letters - Ability to upload GIF images If anyone has any other improvement suggestion please email us or comment here

$$\mathrm{The}\:\mathrm{new}\:\mathrm{symbols}\:\mathrm{are}\:\mathrm{in} \\ $$$$\mathrm{progress}.\:\mathrm{New}\:\mathrm{update}\:\mathrm{will}\:\mathrm{be}\:\mathrm{release} \\ $$$$\mathrm{by}\:\mathrm{end}\:\mathrm{of}\:\mathrm{this}\:\mathrm{month}\:\mathrm{with} \\ $$$$-\:\mathrm{Close}\:\mathrm{integration}\:\mathrm{symbols} \\ $$$$-\:\mathrm{Italic}\:\mathrm{greek}\:\mathrm{capital}\:\mathrm{letters} \\ $$$$-\:\mathrm{Ability}\:\mathrm{to}\:\mathrm{upload}\:\mathrm{GIF}\:\mathrm{images} \\ $$$$ \\ $$$$\mathrm{If}\:\mathrm{anyone}\:\mathrm{has}\:\mathrm{any}\:\mathrm{other}\:\mathrm{improvement} \\ $$$$\mathrm{suggestion}\:\mathrm{please}\:\mathrm{email}\:\mathrm{us}\:\mathrm{or}\:\mathrm{comment} \\ $$$$\mathrm{here} \\ $$

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