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AllQuestion and Answers: Page 190

Question Number 202653    Answers: 0   Comments: 0

Question Number 202651    Answers: 1   Comments: 1

Question Number 202650    Answers: 1   Comments: 0

Question Number 202638    Answers: 1   Comments: 4

(√(x−(1/x)))−(√(1−(1/x)))=1−(1/x)

$$\sqrt{{x}−\frac{\mathrm{1}}{{x}}}−\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}=\mathrm{1}−\frac{\mathrm{1}}{{x}} \\ $$

Question Number 202636    Answers: 0   Comments: 0

((^3 (√4^(5−x) ))/(∫_4 ^6 (x−1)dx)) = (1/2^(2x−1) ) , find the value of x. Solution (4^((5−x)/3) /(∫_4 ^6 ((x^2 /2)−x+k))) = (1/2^(2x−1) ) (2^(2•((5−x)/3)) /(((6^2 /2)−6+k)−((4^2 /2)−4+k))) = (1/2^(2x−1) ) (2^((10−2x)/3) /(((36)/2)−6+k−((16)/2)+4−k)) = (1/2^(2x−1) ) (2^((10−2x)/3) /(18−6−8+4)) = (1/2^(2x−1) ) (2^((10−2x)/3) /8) = (1/2^(2x−1) ) (Cross Multiply) 2^(2x−1) ×2^((10−2x)/3) = 8×1 2^(2x−1) ×2^((10−2x)/3) = 2^3 2^(2x−1+((10−2x)/3)) = 2^3 (Since, the bases are equal. Then, we can equate the exponents) 2x−1+((10−2x)/3) = 3 (Multiply each term by 3) 3(2x)−3(1)+3(((10−2x)/3)) = 3(3) 6x−3+10−2x = 9 (Collect Like Terms) 4x+7 = 9 4x = 9−7 4x = 2 (Divide Both Sides by 4) ((4x)/4) = (2/4) ∴ x = (1/2) soln. by ibroclex_sunshine

$$\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\:^{\mathrm{3}} \sqrt{\mathrm{4}^{\mathrm{5}−\mathrm{x}} }}{\int_{\mathrm{4}} ^{\mathrm{6}} \left(\mathrm{x}−\mathrm{1}\right){dx}}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2x}−\mathrm{1}} }\:,\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underline{\mathrm{Solution}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{4}^{\frac{\mathrm{5}−\mathrm{x}}{\mathrm{3}}} }{\int_{\mathrm{4}} ^{\mathrm{6}} \left(\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{x}+\mathrm{k}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2x}−\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}^{\mathrm{2}\bullet\frac{\mathrm{5}−\mathrm{x}}{\mathrm{3}}} }{\left(\frac{\mathrm{6}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{6}+\mathrm{k}\right)−\left(\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{4}+\mathrm{k}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2x}−\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}^{\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}} }{\frac{\mathrm{36}}{\mathrm{2}}−\mathrm{6}+\cancel{\mathrm{k}}−\frac{\mathrm{16}}{\mathrm{2}}+\mathrm{4}−\cancel{\mathrm{k}}}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2x}−\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}^{\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}} }{\mathrm{18}−\mathrm{6}−\mathrm{8}+\mathrm{4}}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2x}−\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}^{\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}} }{\mathrm{8}}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2x}−\mathrm{1}} }\:\:\left(\mathrm{Cross}\:\mathrm{Multiply}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{2x}−\mathrm{1}} ×\mathrm{2}^{\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}} \:=\:\mathrm{8}×\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{2x}−\mathrm{1}} ×\mathrm{2}^{\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}} \:=\:\mathrm{2}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{2x}−\mathrm{1}+\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}} \:=\:\mathrm{2}^{\mathrm{3}} \:\left(\mathrm{Since},\:\mathrm{the}\:\mathrm{bases}\:\mathrm{are}\:\mathrm{equal}.\:\mathrm{Then},\:\mathrm{we}\:\mathrm{can}\:\mathrm{equate}\:\mathrm{the}\:\mathrm{exponents}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2x}−\mathrm{1}+\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}\:=\:\mathrm{3}\:\left(\mathrm{Multiply}\:\mathrm{each}\:\mathrm{term}\:\mathrm{by}\:\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\left(\mathrm{2x}\right)−\mathrm{3}\left(\mathrm{1}\right)+\cancel{\mathrm{3}}\left(\frac{\mathrm{10}−\mathrm{2x}}{\cancel{\mathrm{3}}}\right)\:=\:\mathrm{3}\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{6x}−\mathrm{3}+\mathrm{10}−\mathrm{2x}\:=\:\mathrm{9}\:\left(\mathrm{Collect}\:\mathrm{Like}\:\mathrm{Terms}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4x}+\mathrm{7}\:=\:\mathrm{9} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4x}\:=\:\mathrm{9}−\mathrm{7} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4x}\:=\:\mathrm{2}\:\left(\mathrm{Divide}\:\mathrm{Both}\:\mathrm{Sides}\:\mathrm{by}\:\mathrm{4}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\cancel{\mathrm{4}x}}{\cancel{\mathrm{4}}}\:=\:\frac{\mathrm{2}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{soln}.\:\mathrm{by}\:\boldsymbol{\mathrm{ibroclex\_sunshine}}\: \\ $$$$ \\ $$

Question Number 202633    Answers: 1   Comments: 0

Question Number 202630    Answers: 1   Comments: 2

Question Number 202616    Answers: 2   Comments: 0

If (a/(a + b)) − ((1 − a)/(a − b)) = x Find ((1 − b)/(a − b)) + (b/(a + b)) = ?

$$\mathrm{If}\:\:\:\:\:\frac{\mathrm{a}}{\mathrm{a}\:+\:\mathrm{b}}\:−\:\frac{\mathrm{1}\:−\:\mathrm{a}}{\mathrm{a}\:−\:\mathrm{b}}\:=\:\mathrm{x} \\ $$$$\mathrm{Find}\:\:\:\:\:\frac{\mathrm{1}\:−\:\mathrm{b}}{\mathrm{a}\:−\:\mathrm{b}}\:+\:\frac{\mathrm{b}}{\mathrm{a}\:+\:\mathrm{b}}\:=\:? \\ $$

Question Number 202614    Answers: 2   Comments: 3

find x

$$\mathrm{find}\:\mathrm{x} \\ $$$$ \\ $$

Question Number 202605    Answers: 1   Comments: 3

Question Number 202604    Answers: 1   Comments: 0

If x = (((√(a + 1)) + (√(a − 1)))/( (√(a + 1)) − (√(a − 1)))) and y = (((√(a + 1)) − (√(a − 1)))/( (√(a + 1)) + (√(a − 1)))) then show that ((x^2 − xy + y^2 )/(x^2 + xy + y^2 )) = ((4a^2 − 3)/(4a^2 − 1)) .

$$\mathrm{If}\:{x}\:=\:\frac{\sqrt{{a}\:+\:\mathrm{1}}\:+\:\sqrt{{a}\:−\:\mathrm{1}}}{\:\sqrt{{a}\:+\:\mathrm{1}}\:−\:\sqrt{{a}\:−\:\mathrm{1}}}\:\mathrm{and}\: \\ $$$${y}\:=\:\frac{\sqrt{{a}\:+\:\mathrm{1}}\:−\:\sqrt{{a}\:−\:\mathrm{1}}}{\:\sqrt{{a}\:+\:\mathrm{1}}\:+\:\sqrt{{a}\:−\:\mathrm{1}}}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$$\frac{{x}^{\mathrm{2}} \:−\:{xy}\:+\:{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} \:+\:{xy}\:+\:{y}^{\mathrm{2}} }\:=\:\frac{\mathrm{4}{a}^{\mathrm{2}} \:−\:\mathrm{3}}{\mathrm{4}{a}^{\mathrm{2}} \:−\:\mathrm{1}}\:. \\ $$

Question Number 202598    Answers: 2   Comments: 0

Question Number 202592    Answers: 0   Comments: 0

If I_n denotes ∫z^n e^(1/z) dz, then show that (n+1)!I_n =I_0 +e^(1/z) (1∙!z^2 +2∙!z^3 +∙∙∙+n!∙z^(n+1) )

$$\:\boldsymbol{{If}}\:\:\boldsymbol{{I}}_{\boldsymbol{{n}}} \:\boldsymbol{{denotes}}\:\int\boldsymbol{{z}}^{\boldsymbol{{n}}} \boldsymbol{{e}}^{\frac{\mathrm{1}}{\boldsymbol{{z}}}} \boldsymbol{{dz}},\:\boldsymbol{{then}}\:\boldsymbol{{show}}\:\boldsymbol{{that}} \\ $$$$\left(\boldsymbol{{n}}+\mathrm{1}\right)!\boldsymbol{{I}}_{\boldsymbol{{n}}} =\boldsymbol{{I}}_{\mathrm{0}} +\boldsymbol{{e}}^{\frac{\mathrm{1}}{\boldsymbol{{z}}}} \left(\mathrm{1}\centerdot!\boldsymbol{{z}}^{\mathrm{2}} +\mathrm{2}\centerdot!\boldsymbol{{z}}^{\mathrm{3}} +\centerdot\centerdot\centerdot+\boldsymbol{{n}}!\centerdot\boldsymbol{{z}}^{\boldsymbol{{n}}+\mathrm{1}} \right) \\ $$$$ \\ $$

Question Number 202591    Answers: 1   Comments: 0

Question Number 202648    Answers: 0   Comments: 5

how to use the fewest “2” to make 2024 you only can use “2” and − + × / ( ) ! ^ ... as the following 2×2×2((2×2)!−2/2)(2×(2+2/2)!−2/2)=2024 totally use thirteen “ 2”

$${how}\:{to}\:{use}\:{the}\:{fewest}\:``\mathrm{2}''\:{to}\:{make}\:\mathrm{2024} \\ $$$${you}\:{only}\:{can}\:{use}\:``\mathrm{2}''\:{and}\:−\:+\:×\:/\:\left(\:\right)\:!\:\:\hat {\:}\:... \\ $$$${as}\:{the}\:{following} \\ $$$$\:\:\:\:\:\:\mathrm{2}×\mathrm{2}×\mathrm{2}\left(\left(\mathrm{2}×\mathrm{2}\right)!−\mathrm{2}/\mathrm{2}\right)\left(\mathrm{2}×\left(\mathrm{2}+\mathrm{2}/\mathrm{2}\right)!−\mathrm{2}/\mathrm{2}\right)=\mathrm{2024} \\ $$$${totally}\:{use}\:{thirteen}\:``\:\mathrm{2}'' \\ $$

Question Number 202584    Answers: 1   Comments: 0

___

$$\:\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \_\_\_ \\ $$

Question Number 202581    Answers: 2   Comments: 1

Question Number 202566    Answers: 2   Comments: 0

Question Number 202589    Answers: 1   Comments: 2

Ω = Σ_(k=1) ^∞ ((1/k) + 2ln(1−(1/(2k))))=?

$$ \\ $$$$\:\:\Omega\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{k}}\:+\:\mathrm{2}{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{k}}\right)\right)=?\:\:\:\:\: \\ $$$$ \\ $$$$ \\ $$

Question Number 202564    Answers: 0   Comments: 0

Find the transformation of Fourier: z(t)=cos(2πF_1 t)+cos(2πF_2 t)

$${Find}\:{the}\:{transformation}\:{of}\:{Fourier}: \\ $$$${z}\left({t}\right)={cos}\left(\mathrm{2}\pi{F}_{\mathrm{1}} {t}\right)+{cos}\left(\mathrm{2}\pi{F}_{\mathrm{2}} {t}\right) \\ $$

Question Number 202551    Answers: 0   Comments: 1

Σ_(i=1) ^n (x+y)_i =(x+y)_1 +(x+y)_2 +...(x+y)_n =x_1 +y_1 +x_2 +y_2 +...+x_n +y_n please it′s correct ?

$$\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{i}} =\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{1}} +\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{2}} +...\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{n}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{x}_{\mathrm{1}} +\mathrm{y}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} +\mathrm{y}_{\mathrm{2}} +...+\mathrm{x}_{\mathrm{n}} +\mathrm{y}_{\mathrm{n}} \: \\ $$$$\mathrm{please}\:\mathrm{it}'\mathrm{s}\:\mathrm{correct}\:? \\ $$$$ \\ $$

Question Number 202543    Answers: 1   Comments: 1

Question Number 202540    Answers: 1   Comments: 0

If (b/(a + b)) = ((3a − b − c)/(2b + c − a)) = ((3c − a)/(2a − b + 3c)) [a + b + c ≠ 0] then show that a = b = c.

$$\mathrm{If}\:\frac{{b}}{{a}\:+\:{b}}\:=\:\frac{\mathrm{3}{a}\:−\:{b}\:−\:{c}}{\mathrm{2}{b}\:+\:{c}\:−\:{a}}\:=\:\frac{\mathrm{3}{c}\:−\:{a}}{\mathrm{2}{a}\:−\:{b}\:+\:\mathrm{3}{c}}\: \\ $$$$\left[{a}\:+\:{b}\:+\:{c}\:\neq\:\mathrm{0}\right]\:\mathrm{then}\:\mathrm{show}\:\mathrm{that}\:{a}\:=\:{b}\:=\:{c}. \\ $$

Question Number 202536    Answers: 2   Comments: 0

f : R→R f(xy)(f(x)−f(y))= (x−y)f(x)f(y)

$$\:\: {f}\::\:\mathrm{R}\rightarrow\mathrm{R}\: \\ $$$$\: {f}\left({xy}\right)\left({f}\left({x}\right)−{f}\left({y}\right)\right)=\:\left({x}−{y}\right){f}\left({x}\right){f}\left({y}\right) \\ $$$$\:\: \\ $$

Question Number 202535    Answers: 1   Comments: 0

Solve for x ((x^3 +3x−3))^(1/3) +((−x^3 −3x+5))^(1/3) =2 (An alteration of Q#202500)

$$\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{{x}} \\ $$$$\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{3}}\:+\sqrt[{\mathrm{3}}]{−{x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{5}}\:=\mathrm{2} \\ $$$$\left(\mathrm{An}\:\mathrm{alteration}\:\mathrm{of}\:\mathrm{Q}#\mathrm{202500}\right) \\ $$

Question Number 202532    Answers: 0   Comments: 1

please help

$${please}\:{help} \\ $$

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