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Question Number 120283 Answers: 0 Comments: 0

fond ∫_2 ^∞ ((ln(1+3x^2 ))/(1+x^4 ))dx

$${fond}\:\int_{\mathrm{2}} ^{\infty} \frac{{ln}\left(\mathrm{1}+\mathrm{3}{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\ $$

Question Number 120275 Answers: 1 Comments: 0

(((3+2 (5)^(1/4) )/(3−2 (5)^(1/4) )))^(1/(4 )) =?

$$\:\sqrt[{\mathrm{4}\:}]{\frac{\mathrm{3}+\mathrm{2}\:\sqrt[{\mathrm{4}}]{\mathrm{5}}}{\mathrm{3}−\mathrm{2}\:\sqrt[{\mathrm{4}}]{\mathrm{5}}}}\:=?\: \\ $$

Question Number 120274 Answers: 0 Comments: 0

Question Number 120272 Answers: 0 Comments: 0

ϕ(x)=x^r where x∈[0,+∞[ and 0<r≤1 shown that for any (x,y)∈R_+ ^2 𝛟(x+y)≤𝛟(x)+𝛟(y) please

$$\varphi\left({x}\right)=\boldsymbol{{x}}^{\boldsymbol{{r}}} \\ $$$$\boldsymbol{{where}}\:\boldsymbol{{x}}\in\left[\mathrm{0},+\infty\left[\:\boldsymbol{{and}}\:\mathrm{0}<\boldsymbol{{r}}\leqslant\mathrm{1}\right.\right. \\ $$$${shown}\:{that}\:{for}\:{any}\:\left(\boldsymbol{{x}},\boldsymbol{{y}}\right)\in\mathbb{R}_{+} ^{\mathrm{2}} \\ $$$$\boldsymbol{\varphi}\left(\boldsymbol{{x}}+\boldsymbol{{y}}\right)\leqslant\boldsymbol{\varphi}\left(\boldsymbol{{x}}\right)+\boldsymbol{\varphi}\left(\boldsymbol{{y}}\right) \\ $$$$\:\:\:\:\boldsymbol{{please}} \\ $$

Question Number 120268 Answers: 3 Comments: 0

Question Number 120258 Answers: 1 Comments: 0

Question Number 120257 Answers: 2 Comments: 1

∫ ((f ′(x))/(f(x))) =?

$$\:\int\:\frac{{f}\:'\left({x}\right)}{{f}\left({x}\right)}\:=? \\ $$

Question Number 120254 Answers: 4 Comments: 0

∫ (dx/(1+cosθ.cos x )) ?

$$\:\int\:\frac{{dx}}{\mathrm{1}+\mathrm{cos}\theta.\mathrm{cos}\:{x}\:}\:? \\ $$

Question Number 120244 Answers: 3 Comments: 2

how to justify that sin (x−((7π)/2) )= cos x

$${how}\:{to}\:{justify}\:\:{that}\:\mathrm{sin}\:\left({x}−\frac{\mathrm{7}\pi}{\mathrm{2}}\:\right)=\:\mathrm{cos}\:{x} \\ $$

Question Number 120243 Answers: 1 Comments: 0

Question Number 120240 Answers: 1 Comments: 0

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Question Number 120241 Answers: 0 Comments: 0

Correction to the last assignment (1)(1/(81^(x−2) )) = 27^(1−x) (1/3^(4(x−2)) )=3^(3(1−x)) (1/3^(4x−8) )=3^(3−3x) 3^(−4x+8) =3^(3−3x) −4x+8=3−3x C.L.T −4x+3x=3−8 −x=−5 x=5 (2)9^x =(1/(729)) 9^x =(1/3^6 ) 3^(2x) =3^(−6) 2x=−6 x=((−6)/2) x=−3 (3)16^x =0.125 2^(4x) =((125)/(1000)) 2^(4x) =(1/8) 2^(4x) =2^(−3) 4x=−3 x=((−3)/4) (4)a^(1/2) =4 a=

$$\boldsymbol{\mathrm{C}}{orrection}\:\boldsymbol{{to}}\:\boldsymbol{{the}}\:\boldsymbol{{last}}\:\boldsymbol{{assignment}} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\frac{\mathrm{1}}{\mathrm{81}^{\boldsymbol{{x}}−\mathrm{2}} }\:=\:\mathrm{27}^{\mathrm{1}−\boldsymbol{{x}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{4}\left(\boldsymbol{{x}}−\mathrm{2}\right)} }=\mathrm{3}^{\mathrm{3}\left(\mathrm{1}−\boldsymbol{{x}}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{4}\boldsymbol{{x}}−\mathrm{8}} }=\mathrm{3}^{\mathrm{3}−\mathrm{3}\boldsymbol{{x}}} \\ $$$$\mathrm{3}^{−\mathrm{4}\boldsymbol{{x}}+\mathrm{8}} =\mathrm{3}^{\mathrm{3}−\mathrm{3}\boldsymbol{{x}}} \\ $$$$−\mathrm{4}\boldsymbol{{x}}+\mathrm{8}=\mathrm{3}−\mathrm{3}\boldsymbol{{x}} \\ $$$$\boldsymbol{{C}}.{L}.{T} \\ $$$$−\mathrm{4}{x}+\mathrm{3}{x}=\mathrm{3}−\mathrm{8} \\ $$$$−{x}=−\mathrm{5} \\ $$$${x}=\mathrm{5} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\mathrm{9}^{{x}} =\frac{\mathrm{1}}{\mathrm{729}} \\ $$$$\mathrm{9}^{{x}} =\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{6}} } \\ $$$$\mathrm{3}^{\mathrm{2}{x}} =\mathrm{3}^{−\mathrm{6}} \\ $$$$\mathrm{2}{x}=−\mathrm{6} \\ $$$${x}=\frac{−\mathrm{6}}{\mathrm{2}} \\ $$$${x}=−\mathrm{3} \\ $$$$ \\ $$$$\left(\mathrm{3}\right)\mathrm{16}^{{x}} =\mathrm{0}.\mathrm{125} \\ $$$$\mathrm{2}^{\mathrm{4}{x}} =\frac{\mathrm{125}}{\mathrm{1000}} \\ $$$$\mathrm{2}^{\mathrm{4}{x}} =\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\mathrm{2}^{\mathrm{4}{x}} =\mathrm{2}^{−\mathrm{3}} \\ $$$$\mathrm{4}{x}=−\mathrm{3} \\ $$$${x}=\frac{−\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$$$\left(\mathrm{4}\right){a}^{\mathrm{1}/\mathrm{2}} =\mathrm{4} \\ $$$${a}= \\ $$

Question Number 120238 Answers: 0 Comments: 0

Question Number 120233 Answers: 0 Comments: 0

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Question Number 120237 Answers: 0 Comments: 1

Question Number 120230 Answers: 3 Comments: 0

∫ (x^3 /((x +2)^4 )) dx OR ∫ ((cos x)/(1 + cos x)) dx a^→ = i^ − j^ + 3k^ and b^(→ ) = 2i^ − 7j^ + k^ .

$$\:\:\:\:\int\:\frac{\mathrm{x}^{\mathrm{3}} }{\left(\mathrm{x}\:+\mathrm{2}\right)^{\mathrm{4}} }\:\mathrm{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{OR} \\ $$$$\:\:\:\int\:\frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{1}\:+\:\mathrm{cos}\:\mathrm{x}}\:\mathrm{dx} \\ $$$$ \\ $$$$\:\:\:\:\:\overset{\rightarrow} {\mathrm{a}}\:=\:\hat {\mathrm{i}}\:−\:\hat {\mathrm{j}}\:+\:\mathrm{3}\hat {\mathrm{k}}\:\:\mathrm{and}\:\overset{\rightarrow\:} {\mathrm{b}}\:=\:\mathrm{2}\hat {\mathrm{i}}\:−\:\mathrm{7}\hat {\mathrm{j}}\:+\:\hat {\mathrm{k}}\:. \\ $$$$ \\ $$$$\:\:\:\: \\ $$

Question Number 120229 Answers: 2 Comments: 0

Question Number 120228 Answers: 0 Comments: 0

Please solve using special function like LambertW function or any other function (if possibe) ((8/7))^x +17^x =25x

$${Please}\:{solve}\:{using}\:{special}\:{function}\:{like}\:{LambertW} \\ $$$${function}\:{or}\:{any}\:{other}\:{function}\:\left({if}\:{possibe}\right) \\ $$$$\left(\frac{\mathrm{8}}{\mathrm{7}}\right)^{{x}} +\mathrm{17}^{{x}} =\mathrm{25}{x} \\ $$

Question Number 120225 Answers: 0 Comments: 0

Question Number 120215 Answers: 1 Comments: 0

Prove that (((((a+b)(b+c)(c+a))/8) ))^(1/3) ≥ (√((ab+bc+ca)/3)) for a,b,c > 0

$${Prove}\:{that}\:\sqrt[{\mathrm{3}}]{\frac{\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)}{\mathrm{8}}\:}\:\geqslant\:\sqrt{\frac{{ab}+{bc}+{ca}}{\mathrm{3}}} \\ $$$${for}\:{a},{b},{c}\:>\:\mathrm{0} \\ $$

Question Number 120211 Answers: 3 Comments: 0

evaluate lim_(x→∞) f(x) and lim_(x→−∞) f(x) for f(x)=(x/( (√(x^2 +1)))).

$$\:{evaluate}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}{f}\left({x}\right)\:{and}\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}{f}\left({x}\right) \\ $$$${for}\:{f}\left({x}\right)=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}. \\ $$

Question Number 120209 Answers: 1 Comments: 0

Peter has 12 relatives (5 man & 7 woman) and his wife also has 12 relatives (5 woman &7 man). They do not have common relatives. They decided to invite 12 guests ,six each of their relatives, such that there are six man and six woman among the guests. How many ways can they choose 12 guests?

$${Peter}\:{has}\:\mathrm{12}\:{relatives}\:\left(\mathrm{5}\:{man}\:\&\:\mathrm{7}\:{woman}\right) \\ $$$${and}\:{his}\:{wife}\:{also}\:{has}\:\mathrm{12}\:{relatives} \\ $$$$\left(\mathrm{5}\:{woman}\:\&\mathrm{7}\:{man}\right).\:{They}\:{do}\:{not} \\ $$$${have}\:{common}\:{relatives}.\:{They}\:{decided} \\ $$$${to}\:{invite}\:\mathrm{12}\:{guests}\:,{six}\:{each}\:{of} \\ $$$${their}\:{relatives},\:{such}\:{that}\:{there} \\ $$$${are}\:{six}\:{man}\:{and}\:{six}\:{woman}\: \\ $$$${among}\:{the}\:{guests}.\:{How}\:{many} \\ $$$${ways}\:{can}\:{they}\:{choose}\:\mathrm{12}\:{guests}? \\ $$

Question Number 120207 Answers: 0 Comments: 0

Question Number 120206 Answers: 0 Comments: 0

Find the remainder when 1×3×5×7×…×2017×2019 is divided by 1000.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when}\: \\ $$$$\mathrm{1}×\mathrm{3}×\mathrm{5}×\mathrm{7}×\ldots×\mathrm{2017}×\mathrm{2019}\:\mathrm{is}\:\mathrm{divided} \\ $$$$\mathrm{by}\:\mathrm{1000}. \\ $$

Question Number 120204 Answers: 0 Comments: 2

Call a 7−digit telephone number d_1 d_2 d_3 −d_4 d_5 d_6 d_7 memorable if the prefix sequence d_1 d_2 d_3 is exactly the same as either of the sequences d_4 d_5 d_6 or d_5 d_6 d_7 (posibly both). Assuming that each d_i can be any of the ten decimal digits 0,1,2,...,9 . Find the number of different memorable telephone numbers

$${Call}\:{a}\:\mathrm{7}−{digit}\:{telephone}\:{number}\:{d}_{\mathrm{1}} {d}_{\mathrm{2}} {d}_{\mathrm{3}} −{d}_{\mathrm{4}} {d}_{\mathrm{5}} {d}_{\mathrm{6}} {d}_{\mathrm{7}} \\ $$$${memorable}\:{if}\:{the}\:{prefix}\:{sequence}\:{d}_{\mathrm{1}} {d}_{\mathrm{2}} {d}_{\mathrm{3}} \\ $$$${is}\:{exactly}\:{the}\:{same}\:{as}\:{either}\:{of}\:{the}\:{sequences} \\ $$$${d}_{\mathrm{4}} {d}_{\mathrm{5}} {d}_{\mathrm{6}} \:{or}\:{d}_{\mathrm{5}} {d}_{\mathrm{6}} {d}_{\mathrm{7}} \:\left({posibly}\:{both}\right). \\ $$$${Assuming}\:{that}\:{each}\:{d}_{{i}} \:{can}\:{be}\:{any}\:{of}\:{the}\:{ten} \\ $$$${decimal}\:{digits}\:\mathrm{0},\mathrm{1},\mathrm{2},...,\mathrm{9}\:.\:{Find}\:{the}\:{number} \\ $$$${of}\:{different}\:{memorable}\:{telephone} \\ $$$${numbers} \\ $$

Question Number 120202 Answers: 0 Comments: 0

Show for the equation a^n = b^2 −1 where n>1 and a>2 are any natural numbers , there are no positive integer solutions for a and b ?

$${Show}\:{for}\:{the}\:{equation}\:{a}^{{n}} \:=\:{b}^{\mathrm{2}} −\mathrm{1}\: \\ $$$${where}\:{n}>\mathrm{1}\:{and}\:{a}>\mathrm{2}\:{are}\:{any}\:{natural} \\ $$$${numbers}\:,\:{there}\:{are}\:{no}\:{positive}\:{integer} \\ $$$${solutions}\:{for}\:{a}\:{and}\:{b}\:? \\ $$

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