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Question Number 68714    Answers: 1   Comments: 0

Question Number 68712    Answers: 0   Comments: 3

given that x and y are two numbers other one. given that a>0 and b>0 and a^x = b^y = (ab)^(xy) show that x + y =0

$${given}\:{that}\:{x}\:{and}\:{y}\:{are}\:{two}\:{numbers}\:{other}\:{one}.\: \\ $$$${given}\:{that}\:\:{a}>\mathrm{0}\:{and}\:{b}>\mathrm{0} \\ $$$${and}\:\:{a}^{{x}} \:=\:{b}^{{y}} \:=\:\left({ab}\right)^{{xy}} \:\:{show}\:{that}\:\:{x}\:+\:{y}\:=\mathrm{0} \\ $$

Question Number 68710    Answers: 0   Comments: 3

Question Number 68703    Answers: 0   Comments: 5

(d/dx)(ln((√((x^2 −1)/(x^2 +1)))))=?

$$\frac{{d}}{{dx}}\left({ln}\left(\sqrt{\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}}\right)\right)=? \\ $$

Question Number 68699    Answers: 1   Comments: 2

∫ ln(x + 4) dx =

$$\int\:{ln}\left({x}\:+\:\mathrm{4}\right)\:{dx}\:= \\ $$

Question Number 68695    Answers: 1   Comments: 0

pour 1<k<n montrer que k(n+1−k)<(n+1/2)^2

$${pour}\:\mathrm{1}<{k}<{n}\:\:\:\:\:{montrer}\:{que} \\ $$$${k}\left({n}+\mathrm{1}−{k}\right)<\left({n}+\mathrm{1}/\mathrm{2}\right)^{\mathrm{2}} \\ $$

Question Number 68693    Answers: 0   Comments: 2

Question Number 68788    Answers: 0   Comments: 1

∫1/(1+x^2 )^n dx

$$\int\mathrm{1}/\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} \:{dx} \\ $$

Question Number 68676    Answers: 0   Comments: 2

Solve the equation tanh^(−1) (((x−2)/(x+1))) = ln 2 show that the set {1,2,4,8} under ×_(15) ,multiplication mod 15 forms a group.

$${Solve}\:{the}\:{equation} \\ $$$${tanh}^{−\mathrm{1}} \left(\frac{{x}−\mathrm{2}}{{x}+\mathrm{1}}\right)\:=\:{ln}\:\mathrm{2} \\ $$$${show}\:{that}\:{the}\:{set}\:\left\{\mathrm{1},\mathrm{2},\mathrm{4},\mathrm{8}\right\}\:\:{under}\:×_{\mathrm{15}} \:,{multiplication}\:{mod}\:\mathrm{15}\:\:{forms}\:{a}\:{group}. \\ $$

Question Number 68675    Answers: 0   Comments: 3

Express in partial fraction f(x) ≡ ((2x^3 + x + 2)/((x^2 +1)(x+1)(x−2))) x ≠ −1,2 Hence or otherwise show that ∫_0 ^1 f(x) dx = −(1/(12))[ 13ln 2 + π]

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:{Express}\:{in}\:{partial}\:{fraction}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:{f}\left({x}\right)\:\equiv\:\frac{\mathrm{2}{x}^{\mathrm{3}} \:+\:{x}\:+\:\mathrm{2}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}+\mathrm{1}\right)\left({x}−\mathrm{2}\right)}\:{x}\:\neq\:−\mathrm{1},\mathrm{2} \\ $$$${Hence}\:{or}\:{otherwise}\:\:{show}\:{that}\:\: \\ $$$$\:\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right)\:{dx}\:=\:−\frac{\mathrm{1}}{\mathrm{12}}\left[\:\mathrm{13}{ln}\:\mathrm{2}\:+\:\pi\right] \\ $$$$ \\ $$

Question Number 68673    Answers: 0   Comments: 8

find sin 20°=?

$${find}\:\boldsymbol{\mathrm{sin}}\:\mathrm{20}°=? \\ $$

Question Number 68664    Answers: 1   Comments: 0

In a equilateral triangle ABC whose side is a, the points M and N are taken on the side BC, such that the triangles ABM, AMN and ANC have the same perimeter. Calculate the distances from vertex A to points M and N. (solve in detail.)

$$\mathrm{In}\:\mathrm{a}\:\mathrm{equilateral}\:\mathrm{triangle}\:{ABC}\:\mathrm{whose} \\ $$$$\mathrm{side}\:\mathrm{is}\:\boldsymbol{{a}},\:\mathrm{the}\:\mathrm{points}\:{M}\:\mathrm{and}\:{N}\:\mathrm{are}\:\mathrm{taken} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{side}\:{BC},\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{triangles} \\ $$$${ABM},\:{AMN}\:\mathrm{and}\:{ANC}\:\mathrm{have}\:\mathrm{the}\:\mathrm{same}\: \\ $$$$\mathrm{perimeter}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{distances}\:\mathrm{from} \\ $$$$\mathrm{vertex}\:{A}\:\mathrm{to}\:\mathrm{points}\:{M}\:\mathrm{and}\:{N}. \\ $$$$\left(\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{detail}}.\right) \\ $$

Question Number 68660    Answers: 0   Comments: 2

∫e^(x+e^x ) dx

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

Question Number 68666    Answers: 1   Comments: 2

Question Number 68642    Answers: 0   Comments: 3

Young′s modulus of a material measures its resistance caused by external stresses. On a vertical wall is a solid mass of specific mass ρ and Young ε modulus in a straight parallelepiped shape, the dimensions of a which are shown in the figure. Based on the correlations between physical quantities, determine the the expression that best represents the deflection suffered by the solid by the action of its own weight.

$$\mathrm{Young}'\mathrm{s}\:\mathrm{modulus}\:\mathrm{of}\:\mathrm{a}\:\mathrm{material}\:\mathrm{measures} \\ $$$$\mathrm{its}\:\mathrm{resistance}\:\mathrm{caused}\:\mathrm{by}\:\mathrm{external}\:\mathrm{stresses}. \\ $$$$\mathrm{On}\:\mathrm{a}\:\mathrm{vertical}\:\mathrm{wall}\:\mathrm{is}\:\mathrm{a}\:\mathrm{solid}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{specific} \\ $$$$\mathrm{mass}\:\rho\:\mathrm{and}\:\mathrm{Young}\:\varepsilon\:\mathrm{modulus}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight} \\ $$$$\mathrm{parallelepiped}\:\mathrm{shape},\:\mathrm{the}\:\mathrm{dimensions} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{which}\:\mathrm{are}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{the}\:\mathrm{figure}.\: \\ $$$$\mathrm{Based}\:\mathrm{on}\:\mathrm{the}\:\mathrm{correlations}\:\mathrm{between}\:\mathrm{physical} \\ $$$$\mathrm{quantities},\:\mathrm{determine}\:\mathrm{the}\:\mathrm{the}\:\mathrm{expression}\:\mathrm{that} \\ $$$$\mathrm{best}\:\mathrm{represents}\:\mathrm{the}\:\mathrm{deflection}\:\mathrm{suffered} \\ $$$$\mathrm{by}\:\mathrm{the}\:\mathrm{solid}\:\mathrm{by}\:\mathrm{the}\:\mathrm{action}\:\mathrm{of}\:\mathrm{its}\:\mathrm{own}\:\mathrm{weight}. \\ $$

Question Number 68636    Answers: 2   Comments: 0

((−a^2 +a+1)/( a^2 +a+1))=((−b^2 +b+1)/( b^2 +b+1)) = (( 2a^2 −2ab+(b−a))/(−2a^2 +2ab+(b−a))) =((−2ab+(a+b)+2)/( 2ab+(a+b)+2)) Solve for a.

$$\frac{−{a}^{\mathrm{2}} +{a}+\mathrm{1}}{\:\:\:\:{a}^{\mathrm{2}} +{a}+\mathrm{1}}=\frac{−{b}^{\mathrm{2}} +{b}+\mathrm{1}}{\:\:\:\:{b}^{\mathrm{2}} +{b}+\mathrm{1}} \\ $$$$\:\:=\:\frac{\:\:\:\:\mathrm{2}{a}^{\mathrm{2}} −\mathrm{2}{ab}+\left({b}−{a}\right)}{−\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{ab}+\left({b}−{a}\right)} \\ $$$$\:\:=\frac{−\mathrm{2}{ab}+\left({a}+{b}\right)+\mathrm{2}}{\:\:\:\:\mathrm{2}{ab}+\left({a}+{b}\right)+\mathrm{2}} \\ $$$${Solve}\:{for}\:\boldsymbol{{a}}. \\ $$$$ \\ $$

Question Number 68629    Answers: 1   Comments: 0

Question Number 68626    Answers: 0   Comments: 0

lim ((−n(2−a)^n )/((2−a))) n→∞

$$\mathrm{l}{im}\:\:\frac{−{n}\left(\mathrm{2}−{a}\right)^{{n}} }{\left(\mathrm{2}−{a}\right)} \\ $$$${n}\rightarrow\infty \\ $$

Question Number 68624    Answers: 2   Comments: 0

Question Number 68618    Answers: 1   Comments: 0

solve for x the following equations a) log x^3 − 2log x^2 + 2log x + 2log (√x) = 3 b) log_x 24 −3log_x 4 + 2log_x 3 =−3

$${solve}\:{for}\:{x}\:{the}\:{following}\:{equations} \\ $$$$\left.{a}\right)\:{log}\:{x}^{\mathrm{3}} \:−\:\mathrm{2}{log}\:{x}^{\mathrm{2}} \:+\:\mathrm{2}{log}\:{x}\:\:+\:\mathrm{2}{log}\:\sqrt{{x}}\:=\:\mathrm{3} \\ $$$$\left.{b}\right)\:{log}_{{x}} \mathrm{24}\:−\mathrm{3}{log}_{{x}} \mathrm{4}\:\:+\:\mathrm{2}{log}_{{x}} \mathrm{3}\:=−\mathrm{3} \\ $$

Question Number 68617    Answers: 0   Comments: 1

find the value of p given that 3^p × 3^(−1) × 5 × 3^(p−1) = 2 × 3^4

$${find}\:{the}\:{value}\:{of}\:{p}\:{given}\:{that} \\ $$$$\mathrm{3}^{{p}} \:×\:\mathrm{3}^{−\mathrm{1}} \:×\:\mathrm{5}\:×\:\mathrm{3}^{{p}−\mathrm{1}} \:=\:\mathrm{2}\:×\:\mathrm{3}^{\mathrm{4}} \\ $$

Question Number 68616    Answers: 2   Comments: 0

given that a,b and c are positive numbers other than 1 , show that log_b a × log_c b × log_a c = 1 hence, evaluate log_(10) 25 × log_2 10 × log_5 4

$${given}\:{that}\:{a},{b}\:{and}\:{c}\:{are}\:{positive}\:{numbers}\:{other}\:{than}\:\mathrm{1} \\ $$$$,\:{show}\:{that}\:\:{log}_{{b}} {a}\:×\:{log}_{{c}} {b}\:×\:{log}_{{a}} {c}\:=\:\mathrm{1} \\ $$$${hence},\:{evaluate}\:\:\:{log}_{\mathrm{10}} \mathrm{25}\:×\:{log}_{\mathrm{2}} \mathrm{10}\:×\:{log}_{\mathrm{5}} \mathrm{4} \\ $$

Question Number 68611    Answers: 1   Comments: 0

Question Number 68609    Answers: 1   Comments: 1

Question Number 68608    Answers: 0   Comments: 0

Question Number 68601    Answers: 0   Comments: 4

ABCD is a side square 1. B, F and E are collinear. FDE is a right triangle with hypotenuse 1 and the DE cathetus is worth x. What is the value of x? (Solve with algebra)

$${ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{side}\:\mathrm{square}\:\mathrm{1}.\: \\ $$$${B},\:{F}\:\mathrm{and}\:{E}\:\mathrm{are}\:\mathrm{collinear}. \\ $$$${FDE}\:\mathrm{is}\:\mathrm{a}\:\mathrm{right}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{hypotenuse}\:\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{the}\:{DE}\:\mathrm{cathetus}\:\mathrm{is}\:\mathrm{worth}\:\boldsymbol{{x}}.\: \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\boldsymbol{{x}}? \\ $$$$\left(\mathrm{Solve}\:\mathrm{with}\:\mathrm{algebra}\right) \\ $$

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