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Question Number 78161 Answers: 1 Comments: 0

Question Number 78160 Answers: 0 Comments: 1

find the term independent of x in [ ((x−1)/x)]^9

$$\mathrm{find}\:\mathrm{the}\:\mathrm{term}\:\mathrm{independent}\:\mathrm{of}\:\mathrm{x}\:\mathrm{in} \\ $$$$\:\:\left[\:\:\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}}\right]^{\mathrm{9}} \\ $$

Question Number 78156 Answers: 0 Comments: 1

Given that u_(n + 1) = (a_n /2) + 5 evalatuate lim_(x→∞) a_n deduce if a_(n ) is convergent or divergent.

$$\mathrm{Given}\:\mathrm{that}\:{u}_{{n}\:+\:\mathrm{1}} =\:\frac{{a}_{{n}} }{\mathrm{2}}\:+\:\mathrm{5}\:\:\: \\ $$$${evalatuate}\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}} \\ $$$${deduce}\:{if}\:{a}_{{n}\:} \:{is}\:{convergent}\:{or}\:{divergent}. \\ $$

Question Number 78075 Answers: 0 Comments: 5

expressing P(x) = ((x^2 + x)/((x−3)(x^2 −2))) in partial fractions gives A. (A/((x−3))) + ((Bx + C)/((x^2 −2))) B. (A/(x−3)) + (B/(x−2)) + (C/(x+2)) C. (A/(x−3)) + (B/(x−(√2))) + (C/(x + (√2))) D. ((Ax + B)/(x−3)) + (C/(x^2 −2))

$${expressing}\:\:{P}\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} \:+\:{x}}{\left({x}−\mathrm{3}\right)\left({x}^{\mathrm{2}} −\mathrm{2}\right)}\:{in}\:{partial}\:{fractions}\:{gives} \\ $$$${A}.\:\:\frac{{A}}{\left({x}−\mathrm{3}\right)}\:+\:\frac{{Bx}\:+\:{C}}{\left({x}^{\mathrm{2}} −\mathrm{2}\right)}\: \\ $$$${B}.\:\:\frac{{A}}{{x}−\mathrm{3}}\:+\:\frac{{B}}{{x}−\mathrm{2}}\:+\:\frac{{C}}{{x}+\mathrm{2}} \\ $$$${C}.\:\frac{{A}}{{x}−\mathrm{3}}\:+\:\frac{{B}}{{x}−\sqrt{\mathrm{2}}}\:+\:\frac{{C}}{{x}\:+\:\sqrt{\mathrm{2}}} \\ $$$${D}.\:\frac{{Ax}\:+\:{B}}{{x}−\mathrm{3}}\:+\:\frac{{C}}{{x}^{\mathrm{2}} −\mathrm{2}} \\ $$

Question Number 78074 Answers: 2 Comments: 0

evaluate ∫_1 ^4 sinh^(−1) x dx and ∫_1 ^(1/2) tanh^(−1) x dx

$${evaluate}\:\int_{\mathrm{1}} ^{\mathrm{4}} \mathrm{sinh}\:^{−\mathrm{1}} {x}\:{dx}\:\:{and}\:\underset{\mathrm{1}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}\mathrm{tanh}\:^{−\mathrm{1}} {x}\:{dx} \\ $$

Question Number 77872 Answers: 2 Comments: 6

show that f(x)=2r^3 +5x−1 has a zero in the interval [0.1].

$${show}\:{that}\:{f}\left({x}\right)=\mathrm{2}{r}^{\mathrm{3}} +\mathrm{5}{x}−\mathrm{1}\:{has}\:{a}\:{zero}\:{in}\:{the}\:{interval}\:\left[\mathrm{0}.\mathrm{1}\right]. \\ $$

Question Number 77845 Answers: 0 Comments: 0

In the equation B=μ_0 H×μ_0 M why is the polarization of the vacuum accounted for by constant μ_0 if the vacuum is absolutely empty?

$$\boldsymbol{\mathrm{I}}\mathrm{n}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\:\boldsymbol{\mathrm{B}}=\mu_{\mathrm{0}} \boldsymbol{\mathrm{H}}×\mu_{\mathrm{0}} \boldsymbol{\mathrm{M}}\: \\ $$$$\mathrm{why}\:\mathrm{is}\:\mathrm{the}\:\mathrm{polarization}\:\mathrm{of}\:\mathrm{the}\:\mathrm{vacuum}\: \\ $$$$\mathrm{accounted}\:\mathrm{for}\:\mathrm{by}\: \\ $$$$\mathrm{constant}\:\mu_{\mathrm{0}} \\ $$$$\mathrm{if}\:\mathrm{the}\:\mathrm{vacuum}\:\mathrm{is}\:\mathrm{absolutely}\:\mathrm{empty}? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 77803 Answers: 1 Comments: 0

solve (4x^2 +4x+1)y′′−(12x+6)y′−8x^3 −1=12x^2 −16y+6x

$${solve} \\ $$$$\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}\right){y}''−\left(\mathrm{12}{x}+\mathrm{6}\right){y}'−\mathrm{8}{x}^{\mathrm{3}} −\mathrm{1}=\mathrm{12}{x}^{\mathrm{2}} −\mathrm{16}{y}+\mathrm{6}{x} \\ $$

Question Number 77802 Answers: 0 Comments: 0

Question Number 77772 Answers: 0 Comments: 3

Question Number 77722 Answers: 1 Comments: 2

how to find n−term from S_n =n^2 +7n+2 ?

$${how}\:{to}\:{find}\: \\ $$$${n}−{term}\:{from} \\ $$$${S}_{{n}} ={n}^{\mathrm{2}} +\mathrm{7}{n}+\mathrm{2}\:? \\ $$

Question Number 77631 Answers: 1 Comments: 0

x^(log_3 (2)) =(√x)+1

$${x}^{{log}_{\mathrm{3}} \left(\mathrm{2}\right)} =\sqrt{{x}}+\mathrm{1} \\ $$

Question Number 77573 Answers: 0 Comments: 2

xln x =((143851)/(40000)) solve for x ⇒ nice surprise

$${x}\mathrm{ln}\:{x}\:=\frac{\mathrm{143851}}{\mathrm{40000}} \\ $$$$\mathrm{solve}\:\mathrm{for}\:{x}\:\Rightarrow\:\mathrm{nice}\:\mathrm{surprise} \\ $$

Question Number 77336 Answers: 0 Comments: 2

make x subject of formula x^y^x + 8x = y

$${make}\:\boldsymbol{{x}}\:{subject}\:{of}\:{formula} \\ $$$$ \\ $$$$\boldsymbol{{x}}^{\boldsymbol{{y}}^{\boldsymbol{{x}}} } \:+\:\mathrm{8}\boldsymbol{{x}}\:\:=\:\:\boldsymbol{{y}} \\ $$

Question Number 77186 Answers: 1 Comments: 0

given { ((3^y −1= (6/2^x ))),(((3)^(y/x) = 2 )) :} find (1/x)+(1/y).

$$\mathrm{given}\: \\ $$$$\begin{cases}{\mathrm{3}^{\mathrm{y}} −\mathrm{1}=\:\frac{\mathrm{6}}{\mathrm{2}^{\mathrm{x}} }}\\{\left(\mathrm{3}\right)^{\frac{\mathrm{y}}{\mathrm{x}}} \:=\:\mathrm{2}\:}\end{cases}\:\:\mathrm{find}\:\frac{\mathrm{1}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{y}}. \\ $$

Question Number 77180 Answers: 1 Comments: 0

given a quadratic equation 3x^2 −x+(t^2 −4t+3)=0 has roots sin α and cos α. find the value (√(t^2 −4t+5)) .

$$ \\ $$$$ \\ $$$$\mathrm{given}\:\mathrm{a}\:\mathrm{quadratic}\:\mathrm{equation}\: \\ $$$$\mathrm{3x}^{\mathrm{2}} −\mathrm{x}+\left(\mathrm{t}^{\mathrm{2}} −\mathrm{4t}+\mathrm{3}\right)=\mathrm{0}\:\mathrm{has} \\ $$$$\mathrm{roots}\:\mathrm{sin}\:\alpha\:\mathrm{and}\:\mathrm{cos}\:\alpha.\:\mathrm{find}\:\mathrm{the}\: \\ $$$$\mathrm{value}\:\sqrt{\mathrm{t}^{\mathrm{2}} −\mathrm{4t}+\mathrm{5}}\:. \\ $$

Question Number 77160 Answers: 1 Comments: 1

Question Number 77149 Answers: 0 Comments: 2

Any reference to a book or video that coould help me solve Differential equations? please help

$$\mathrm{Any}\:\mathrm{reference}\:\mathrm{to}\:\mathrm{a}\:\mathrm{book}\:\mathrm{or}\:\mathrm{video} \\ $$$$\mathrm{that}\:\mathrm{coould}\:\mathrm{help}\:\mathrm{me}\:\mathrm{solve}\:\mathrm{Differential}\:\mathrm{equations}?\: \\ $$$$\mathrm{please}\:\mathrm{help} \\ $$

Question Number 77147 Answers: 0 Comments: 1

Σ_(r=1) ^∞ (1/r^k ) is divergent for: A. k ≤ 1 B. k > 2 C. k ≤ 2 D. 0 ≤ k < 2

$$\:\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{r}^{{k}} }\:{is}\:{divergent}\:{for}: \\ $$$${A}.\:{k}\:\leqslant\:\mathrm{1} \\ $$$${B}.\:{k}\:>\:\mathrm{2} \\ $$$${C}.\:{k}\:\leqslant\:\mathrm{2} \\ $$$${D}.\:\mathrm{0}\:\leqslant\:{k}\:<\:\mathrm{2} \\ $$

Question Number 76973 Answers: 2 Comments: 0

In a ABC triangle the side a=6 and c^2 −b^2 =66. Calculate the projections of sides b and c on a.

$${In}\:{a}\:{ABC}\:{triangle}\:{the}\:{side}\:\boldsymbol{{a}}=\mathrm{6}\:{and} \\ $$$$\boldsymbol{{c}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} =\mathrm{66}.\:{Calculate}\:{the}\:{projections} \\ $$$${of}\:{sides}\:\boldsymbol{{b}}\:{and}\:\boldsymbol{{c}}\:{on}\:\boldsymbol{{a}}. \\ $$

Question Number 76922 Answers: 0 Comments: 1

Question Number 76819 Answers: 0 Comments: 4

one of the foci of the ellipse (x^2 /9) + (y^2 /4) = 1 is A. (4,0) B. (9,0) C. (5,0) D. ((√5) , 0)

$$\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{foci}\:\mathrm{of}\:\mathrm{the}\:\mathrm{ellipse} \\ $$$$\:\:\:\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{9}}\:+\:\frac{{y}^{\mathrm{2}} }{\mathrm{4}}\:=\:\mathrm{1}\:\mathrm{is} \\ $$$$\mathrm{A}.\:\left(\mathrm{4},\mathrm{0}\right) \\ $$$$\mathrm{B}.\:\left(\mathrm{9},\mathrm{0}\right) \\ $$$$\mathrm{C}.\:\left(\mathrm{5},\mathrm{0}\right) \\ $$$$\mathrm{D}.\:\left(\sqrt{\mathrm{5}}\:,\:\mathrm{0}\right) \\ $$

Question Number 76817 Answers: 0 Comments: 2

A compound pendulum ocsillates through angles θ about its equilibrium position such that 8aθ^2 = 9g cosθ, a>0. its period is A. 2π(√((8a)/(9g))) B. ((3π)/8)(√(a/g)) C. 2π(√((9g)/(8a))) D. ((8π)/3)(√(a/g))

$$\mathrm{A}\:\mathrm{compound}\:\mathrm{pendulum}\:\mathrm{ocsillates} \\ $$$$\mathrm{through}\:\mathrm{angles}\:\theta\:\mathrm{about}\:\mathrm{its}\:\mathrm{equilibrium} \\ $$$$\mathrm{position}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\mathrm{8}{a}\theta^{\mathrm{2}} \:=\:\mathrm{9}{g}\:{cos}\theta,\:{a}>\mathrm{0}.\:\mathrm{its}\:\mathrm{period}\:\mathrm{is}\: \\ $$$$\mathrm{A}.\:\mathrm{2}\pi\sqrt{\frac{\mathrm{8}{a}}{\mathrm{9}{g}}} \\ $$$$\mathrm{B}.\:\frac{\mathrm{3}\pi}{\mathrm{8}}\sqrt{\frac{{a}}{{g}}} \\ $$$$\mathrm{C}.\:\mathrm{2}\pi\sqrt{\frac{\mathrm{9}{g}}{\mathrm{8}{a}}} \\ $$$$\mathrm{D}.\:\frac{\mathrm{8}\pi}{\mathrm{3}}\sqrt{\frac{{a}}{{g}}} \\ $$

Question Number 76813 Answers: 1 Comments: 7

Σ_(k=1) ^(2n) (−1)^k = A. ∞ B. 1 C. −1 D. 0

$$\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{2n}} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \:=\: \\ $$$$\mathrm{A}.\:\infty \\ $$$$\mathrm{B}.\:\mathrm{1} \\ $$$$\mathrm{C}.\:−\mathrm{1} \\ $$$$\mathrm{D}.\:\mathrm{0} \\ $$

Question Number 76811 Answers: 0 Comments: 2

The eccentricity of the hyperbola (x^2 /(64)) − (y^2 /(36)) = 1 is A. (5/4) B. (3/4) C. (4/5) D. (4/3)

$$\mathrm{The}\:\mathrm{eccentricity}\:\mathrm{of}\:\mathrm{the}\:\mathrm{hyperbola} \\ $$$$\:\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{64}}\:−\:\frac{{y}^{\mathrm{2}} }{\mathrm{36}}\:=\:\mathrm{1}\:\mathrm{is}\: \\ $$$$\mathrm{A}.\:\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\mathrm{B}.\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{C}.\:\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\mathrm{D}.\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$

Question Number 76809 Answers: 0 Comments: 3

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