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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)

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

Question Number 68600 Answers: 0 Comments: 1

1)calculatef(a)= ∫_0 ^∞ ((cos(arctanx))/(a+x^2 ))dx with a>0 2)?calculste g(a) =∫_0 ^∞ ((cos(arctanx))/((a+x^2 )^2 )) 3)find the value if integrals ∫_0 ^∞ ((cos(arctanx))/(2+x^2 )) and ∫_0 ^∞ ((cos(arctanx))/((1+x^2 )^2 ))

$$\left.\mathrm{1}\right){calculatef}\left({a}\right)=\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({arctanx}\right)}{{a}+{x}^{\mathrm{2}} }{dx}\:{with}\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right)?{calculste}\:\:{g}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left({arctanx}\right)}{\left({a}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right){find}\:{the}\:{value}\:{if}\:{integrals} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({arctanx}\right)}{\mathrm{2}+{x}^{\mathrm{2}} }\:{and}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({arctanx}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$

Question Number 68598 Answers: 0 Comments: 2

find ∫ (dx/(x^3 −4x +3))

$$\:{find}\:\int\:\:\:\:\frac{{dx}}{{x}^{\mathrm{3}} −\mathrm{4}{x}\:+\mathrm{3}} \\ $$

Question Number 68597 Answers: 0 Comments: 1

calculate ∫_0 ^∞ ((arctan(e^x^2 ))/(x^2 +8))dx

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left({e}^{{x}^{\mathrm{2}} } \right)}{{x}^{\mathrm{2}} \:+\mathrm{8}}{dx} \\ $$

Question Number 68596 Answers: 0 Comments: 1

calculate ∫_(π/2) ^(π/3) ((xdx)/(3+cosx))

$${calculate}\:\int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{3}}} \:\:\:\:\frac{{xdx}}{\mathrm{3}+{cosx}} \\ $$

Question Number 68595 Answers: 0 Comments: 1

calculate A_λ =∫_0 ^∞ (e^(−λx^2 ) /(x^4 +1))dx with λ>0 and find ∫_0 ^1 A_λ dλ