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Question Number 144170 Answers: 2 Comments: 0

Question Number 144169 Answers: 1 Comments: 1

please find the value of sin ((2Π)/7)+sin ((4Π)/7)+sin ((8Π)/7)

$$\mathrm{please}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\mathrm{sin}\:\frac{\mathrm{2}\Pi}{\mathrm{7}}+\mathrm{sin}\:\frac{\mathrm{4}\Pi}{\mathrm{7}}+\mathrm{sin}\:\frac{\mathrm{8}\Pi}{\mathrm{7}} \\ $$

Question Number 144168 Answers: 1 Comments: 0

Question Number 144166 Answers: 1 Comments: 0

Given p^→ =(√2) i^ +2(√3) j^ + (√3) k^ & q^→ =a i^ +j^ +2k^ . If proj_q^→ p^→ = ((2(√2))/9) q^→ then ∣q^→ ∣ =?

$$\mathrm{Given}\:\overset{\rightarrow} {\mathrm{p}}=\sqrt{\mathrm{2}}\:\hat {\mathrm{i}}+\mathrm{2}\sqrt{\mathrm{3}}\:\hat {\mathrm{j}}+\:\sqrt{\mathrm{3}}\:\hat {\mathrm{k}}\:\&\: \\ $$$$\:\overset{\rightarrow} {\mathrm{q}}=\mathrm{a}\:\hat {\mathrm{i}}+\hat {\mathrm{j}}\:+\mathrm{2}\hat {\mathrm{k}}\:.\:\mathrm{If}\:\mathrm{proj}_{\overset{\rightarrow} {\mathrm{q}}} \:\overset{\rightarrow} {\mathrm{p}}\:=\:\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{9}}\:\overset{\rightarrow} {\mathrm{q}}\: \\ $$$$\mathrm{then}\:\mid\overset{\rightarrow} {\mathrm{q}}\mid\:=?\: \\ $$

Question Number 144164 Answers: 1 Comments: 0

In ΔABC given a=5, b=7 & c= 4. If ∡CAB = α then cot ((1/2)α)=?

$$\mathrm{In}\:\Delta\mathrm{ABC}\:\mathrm{given}\:\mathrm{a}=\mathrm{5},\:\mathrm{b}=\mathrm{7}\:\&\:\mathrm{c}=\:\mathrm{4}. \\ $$$$\mathrm{If}\:\measuredangle\mathrm{CAB}\:=\:\alpha\:\mathrm{then}\:\mathrm{cot}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\alpha\right)=? \\ $$

Question Number 144162 Answers: 0 Comments: 1

Question Number 144156 Answers: 1 Comments: 2

Question Number 144152 Answers: 2 Comments: 0

I=∫((e^x^2 +e^x )/(e^x^2 +1))dx=?

$$\mathrm{I}=\int\frac{\mathrm{e}^{\mathrm{x}^{\mathrm{2}} } +\mathrm{e}^{\mathrm{x}} }{\mathrm{e}^{\mathrm{x}^{\mathrm{2}} } +\mathrm{1}}\mathrm{dx}=? \\ $$

Question Number 144148 Answers: 1 Comments: 0

{ ((2ln x+ln y = 2)),((x^2 + y = e^2 +1)) :}

$$\begin{cases}{\mathrm{2ln}\:\mathrm{x}+\mathrm{ln}\:\mathrm{y}\:=\:\mathrm{2}}\\{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}\:=\:\mathrm{e}^{\mathrm{2}} +\mathrm{1}}\end{cases} \\ $$

Question Number 144143 Answers: 1 Comments: 0

∫_(1/a) ^a ((arctg(x))/x)dx=???

$$\int_{\frac{\mathrm{1}}{{a}}} ^{{a}} \frac{{arctg}\left({x}\right)}{{x}}{dx}=??? \\ $$

Question Number 144142 Answers: 1 Comments: 0

∫ ((√(cos x+(√(cos x+(√(cos x+(√(cos x+(√(...))))))))))/(sin x)) dx

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

Question Number 144139 Answers: 2 Comments: 0

Question Number 144136 Answers: 0 Comments: 1

Question Number 144120 Answers: 2 Comments: 0

log_2 3 = x , log_3 5 = y , lg6 = ?

$${log}_{\mathrm{2}} \mathrm{3}\:=\:{x}\:,\:{log}_{\mathrm{3}} \mathrm{5}\:=\:{y}\:,\:{lg}\mathrm{6}\:=\:? \\ $$

Question Number 144118 Answers: 0 Comments: 0

Let a,b,c > 0 and a+b+c = 3. Prove that (((√(ab))+1)/( (√(ab))+(√c)))+(((√(bc))+1)/( (√(bc))+(√a)))+(((√(ca))+1)/( (√(ca))+(√b))) ≥ (√a)+(√b)+(√c)

$$\mathrm{Let}\:{a},{b},{c}\:>\:\mathrm{0}\:\mathrm{and}\:{a}+{b}+{c}\:=\:\mathrm{3}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\sqrt{{ab}}+\mathrm{1}}{\:\sqrt{{ab}}+\sqrt{{c}}}+\frac{\sqrt{{bc}}+\mathrm{1}}{\:\sqrt{{bc}}+\sqrt{{a}}}+\frac{\sqrt{{ca}}+\mathrm{1}}{\:\sqrt{{ca}}+\sqrt{{b}}}\:\geqslant\:\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

Question Number 144116 Answers: 0 Comments: 0

Question Number 144109 Answers: 3 Comments: 1

hi, everybody ! 1. calculate : I =∫_(𝛑/6) ^( (𝛑/3)) ln(tan x)dx. 2. calculate : lim_(x → e) ((x(√(1−ln x)))/(x−e)) .

$$\boldsymbol{\mathrm{hi}},\:\boldsymbol{\mathrm{everybody}}\:! \\ $$$$\mathrm{1}.\:\boldsymbol{\mathrm{calculate}}\::\:\boldsymbol{\mathrm{I}}\:=\int_{\frac{\boldsymbol{\pi}}{\mathrm{6}}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{3}}} \boldsymbol{{ln}}\left(\boldsymbol{{tan}}\:\boldsymbol{{x}}\right)\boldsymbol{{dx}}. \\ $$$$\mathrm{2}.\:\boldsymbol{\mathrm{calculate}}\:\::\:\underset{\boldsymbol{{x}}\:\rightarrow\:\boldsymbol{{e}}} {\boldsymbol{{lim}}}\:\frac{\boldsymbol{{x}}\sqrt{\mathrm{1}−\boldsymbol{{ln}}\:\boldsymbol{{x}}}}{\boldsymbol{{x}}−\boldsymbol{{e}}}\:. \\ $$

Question Number 144107 Answers: 1 Comments: 0

A=lim_(n→+∝) ((1+(2)^(1/7) +(3)^(1/7) +(4)^(1/7) +.....+(n)^(1/7) )/( (n^9 )^(1/7) )) =?

$$\mathrm{A}=\underset{\mathrm{n}\rightarrow+\propto} {\mathrm{lim}}\frac{\mathrm{1}+\sqrt[{\mathrm{7}}]{\mathrm{2}}+\sqrt[{\mathrm{7}}]{\mathrm{3}}+\sqrt[{\mathrm{7}}]{\mathrm{4}}+.....+\sqrt[{\mathrm{7}}]{\mathrm{n}}}{\:\sqrt[{\mathrm{7}}]{\mathrm{n}^{\mathrm{9}} }}\:=? \\ $$

Question Number 144106 Answers: 1 Comments: 1

lim_(x→∞) ((4^(x-2) + 3^x + 2^x )/(4^(x-1) + 3^(x+1) )) = ?

$$\underset{{x}\rightarrow\infty} {{lim}}\frac{\mathrm{4}^{{x}-\mathrm{2}} \:+\:\mathrm{3}^{{x}} \:+\:\mathrm{2}^{{x}} }{\mathrm{4}^{{x}-\mathrm{1}} \:+\:\mathrm{3}^{{x}+\mathrm{1}} }\:=\:? \\ $$

Question Number 144095 Answers: 1 Comments: 0

if x>0 , r=pq , 1≤p≤q then: 1+rx ≤ (1+qx)^p ≤ (1+px)^q ≤ (1+x)^r

$${if}\:\:{x}>\mathrm{0}\:,\:{r}={pq}\:,\:\mathrm{1}\leqslant{p}\leqslant{q}\:\:{then}: \\ $$$$\mathrm{1}+{rx}\:\leqslant\:\left(\mathrm{1}+{qx}\right)^{\boldsymbol{{p}}} \:\leqslant\:\left(\mathrm{1}+{px}\right)^{\boldsymbol{{q}}} \:\leqslant\:\left(\mathrm{1}+{x}\right)^{\boldsymbol{{r}}} \\ $$

Question Number 144094 Answers: 1 Comments: 1

Question Number 144083 Answers: 1 Comments: 0

if a;b;c>0 then: ((9abc)/(a^3 + b^3 + c^3 )) + (a^2 /(bc)) + (b^2 /(ca)) + (c^2 /(ab)) ≥ 6

$${if}\:\:{a};{b};{c}>\mathrm{0}\:\:{then}: \\ $$$$\frac{\mathrm{9}{abc}}{{a}^{\mathrm{3}} \:+\:{b}^{\mathrm{3}} \:+\:{c}^{\mathrm{3}} }\:+\:\frac{{a}^{\mathrm{2}} }{{bc}}\:+\:\frac{{b}^{\mathrm{2}} }{{ca}}\:+\:\frac{{c}^{\mathrm{2}} }{{ab}}\:\geqslant\:\mathrm{6} \\ $$

Question Number 144074 Answers: 1 Comments: 0

The parallelogram ABCD has ∣∣AB∣∣ =6, ∣∣AC∣∣=7 & d(D,AC)=2 Find d(D,AB).

$$\:\mathrm{The}\:\mathrm{parallelogram}\:\mathrm{ABCD}\:\mathrm{has} \\ $$$$\mid\mid\mathrm{AB}\mid\mid\:=\mathrm{6},\:\mid\mid\mathrm{AC}\mid\mid=\mathrm{7}\:\&\:\mathrm{d}\left(\mathrm{D},\mathrm{AC}\right)=\mathrm{2} \\ $$$$\mathrm{Find}\:\mathrm{d}\left(\mathrm{D},\mathrm{AB}\right). \\ $$

Question Number 144072 Answers: 2 Comments: 0

Use the reduction formula to rewrite −3 sin x −3 cos x in the form K sin (x+α) .

$$\mathrm{Use}\:\mathrm{the}\:\mathrm{reduction}\:\mathrm{formula}\:\mathrm{to} \\ $$$$\mathrm{rewrite}\:−\mathrm{3}\:\mathrm{sin}\:\mathrm{x}\:−\mathrm{3}\:\mathrm{cos}\:\mathrm{x}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{form}\:\mathrm{K}\:\mathrm{sin}\:\left(\mathrm{x}+\alpha\right)\:. \\ $$

Question Number 144069 Answers: 0 Comments: 0

I=∫_0 ^(π/2) ((2304cosx)/((cos4x−8cos2x+15)^2 ))dx=?

$$\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{2304cosx}}{\left(\mathrm{cos4x}−\mathrm{8cos2x}+\mathrm{15}\right)^{\mathrm{2}} }\mathrm{dx}=? \\ $$

Question Number 144068 Answers: 1 Comments: 0

S_(2019) =1+ (1/2^2 ) + (1/3^2 ) + ...+ (1/(2019^2 )) =?

$$\mathrm{S}_{\mathrm{2019}} =\mathrm{1}+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+\:...+\:\frac{\mathrm{1}}{\mathrm{2019}^{\mathrm{2}} }\:=? \\ $$

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