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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

$${calculate}\:\:{A}_{\lambda} =\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{e}^{−\lambda{x}^{\mathrm{2}} } }{{x}^{\mathrm{4}} +\mathrm{1}}{dx}\:\:{with}\:\lambda>\mathrm{0}\:\:{and}\: \\ $$$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{A}_{\lambda} \:{d}\lambda \\ $$

Question Number 68594    Answers: 0   Comments: 0

calculate ∫_0 ^(2π) (dx/(cosx +sin(2x)))

$${calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dx}}{{cosx}\:+{sin}\left(\mathrm{2}{x}\right)} \\ $$

Question Number 68593    Answers: 0   Comments: 1

calculate Σ_(n=1) ^∞ (((−1)^n )/(n(2n+1)^2 ))

$${calculate}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Question Number 68592    Answers: 0   Comments: 2

find Σ_(n=1) ^∞ (1/(n^2 (n+1)^3 ))

$${find}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$

Question Number 68591    Answers: 0   Comments: 6

Question Number 68589    Answers: 1   Comments: 0

Question Number 68632    Answers: 0   Comments: 0

(sin(π/9) + i sin((3π)/(18)))^(−9)

$$\left({sin}\frac{\pi}{\mathrm{9}}\:+\:{i}\:{sin}\frac{\mathrm{3}\pi}{\mathrm{18}}\right)^{−\mathrm{9}} \\ $$

Question Number 68554    Answers: 0   Comments: 4

Question Number 68549    Answers: 1   Comments: 3

y=(x^3 /(x^2 +1)) y^(−1) =...

$${y}=\frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$${y}^{−\mathrm{1}} =... \\ $$

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