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AlgebraQuestion and Answers: Page 91

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Question Number 182108 Answers: 2 Comments: 0

A truck, P traveling at 54km/hr passes through a point at 10:30am, while another truck, Q traveling at 90km/hr passes through this same point 30 minutes later. At what time will truck Q overtake P?

Question Number 182093 Answers: 1 Comments: 2

Solve the equation: ((x−6)/(2020))+((x−5)/(2021))+((x−4)/(2022))=3

$${Solve}\:{the}\:{equation}: \\ $$$$\frac{{x}−\mathrm{6}}{\mathrm{2020}}+\frac{{x}−\mathrm{5}}{\mathrm{2021}}+\frac{{x}−\mathrm{4}}{\mathrm{2022}}=\mathrm{3} \\ $$

Question Number 182074 Answers: 2 Comments: 0

Let x+ xy+ y= 54 ; x, y∈ N , Find x+ y

$${Let}\:{x}+\:{xy}+\:{y}=\:\mathrm{54}\:\:\:;\:{x},\:{y}\in\:\mathbb{N}\:,\:{Find}\:{x}+\:{y} \\ $$

Question Number 182073 Answers: 1 Comments: 0

Find the sum of the solutions of the equation: ∣(√x) − 2∣+ (√x) ((√x) − 4)+ 2= 0 ; x> 0

$${Find}\:{the}\:{sum}\:{of}\:{the}\:{solutions}\:{of}\:{the}\:{equation}: \\ $$$$\:\mid\sqrt{{x}}\:−\:\mathrm{2}\mid+\:\sqrt{{x}}\:\left(\sqrt{{x}}\:−\:\mathrm{4}\right)+\:\mathrm{2}=\:\mathrm{0}\:\:\:;\:{x}>\:\mathrm{0} \\ $$

Question Number 182043 Answers: 0 Comments: 0

For a, b, c, d ∈ R a+b+c+d=0 ab, ac, ad, bc, bd, cd ≠0 Prove the inequality: ((ab)/((a+b)^2 ))+((ac)/((a+c)^2 ))+((ad)/((a+d)^2 ))+((bc)/((b+c)^2 ))+((bd)/((b+d)^2 ))+((cd)/((c+d)^2 ))≤−(3/2) When is equality reached?

$${For}\:{a},\:{b},\:{c},\:{d}\:\in\:\mathbb{R} \\ $$$${a}+{b}+{c}+{d}=\mathrm{0} \\ $$$${ab},\:{ac},\:{ad},\:{bc},\:{bd},\:{cd}\:\neq\mathrm{0} \\ $$$${Prove}\:{the}\:{inequality}: \\ $$$$\frac{{ab}}{\left({a}+{b}\right)^{\mathrm{2}} }+\frac{{ac}}{\left({a}+{c}\right)^{\mathrm{2}} }+\frac{{ad}}{\left({a}+{d}\right)^{\mathrm{2}} }+\frac{{bc}}{\left({b}+{c}\right)^{\mathrm{2}} }+\frac{{bd}}{\left({b}+{d}\right)^{\mathrm{2}} }+\frac{{cd}}{\left({c}+{d}\right)^{\mathrm{2}} }\leq−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${When}\:{is}\:{equality}\:{reached}? \\ $$

Question Number 182012 Answers: 1 Comments: 0

f(x)=9^x −m∙3^x +m+6 ∃x∈R, f(x)+f(−x)=0 find the range of m.

$${f}\left({x}\right)=\mathrm{9}^{{x}} −{m}\centerdot\mathrm{3}^{{x}} +{m}+\mathrm{6} \\ $$$$\exists{x}\in\mathbb{R},\:{f}\left({x}\right)+{f}\left(−{x}\right)=\mathrm{0} \\ $$$$\mathrm{find}\:\mathrm{the}\:\:\mathrm{range}\:\mathrm{of}\:{m}. \\ $$

Question Number 182000 Answers: 0 Comments: 1

if a−2b+3c−4d+5e−6f=0, find the maximum of ((∣a+b+c+d+e+f∣)/( (√(a^2 +b^2 +c^2 +d^2 +e^2 +f^2 )))).

$${if}\:\boldsymbol{{a}}−\mathrm{2}\boldsymbol{{b}}+\mathrm{3}\boldsymbol{{c}}−\mathrm{4}\boldsymbol{{d}}+\mathrm{5}\boldsymbol{{e}}−\mathrm{6}\boldsymbol{{f}}=\mathrm{0},\:{find} \\ $$$${the}\:{maximum}\:{of} \\ $$$$\frac{\mid\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}+\boldsymbol{{d}}+\boldsymbol{{e}}+\boldsymbol{{f}}\mid}{\:\sqrt{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} +\boldsymbol{{d}}^{\mathrm{2}} +\boldsymbol{{e}}^{\mathrm{2}} +\boldsymbol{{f}}^{\mathrm{2}} }}. \\ $$

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f(x)=2^x +3^x −6^x Find f(x)_(max)

$${f}\left({x}\right)=\mathrm{2}^{{x}} +\mathrm{3}^{{x}} −\mathrm{6}^{{x}} \\ $$$$\mathrm{Find}\:{f}\left({x}\right)_{\mathrm{max}} \\ $$

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help ! ∫ ((ln(x+1))/(x^2 +1))dx = ???

$$\mathrm{help}\:! \\ $$$$\int\:\frac{{ln}\left({x}+\mathrm{1}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}\:=\:??? \\ $$

Question Number 181837 Answers: 3 Comments: 2

find integers a>b>c>0 such that (1/a)+(2/b)+(3/c)=1

$${find}\:{integers}\:{a}>{b}>{c}>\mathrm{0}\:{such}\:{that} \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{2}}{{b}}+\frac{\mathrm{3}}{{c}}=\mathrm{1} \\ $$

Question Number 181794 Answers: 1 Comments: 0

f(0) = 0 f(1) = e^3 f ∈ C^2 (R) f^(′′) (x) − 5 f^′ (x) + 6 f(x) = 0 , ∀ x ∈ R Find: 𝛀 =lim_(x→∞) (1 − (1/(f (x))))^x

$$\mathrm{f}\left(\mathrm{0}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{f}\left(\mathrm{1}\right)\:=\:\mathrm{e}^{\mathrm{3}} \\ $$$$\mathrm{f}\:\in\:\mathbb{C}^{\mathrm{2}} \:\left(\mathbb{R}\right) \\ $$$$\mathrm{f}\:^{''} \:\left(\mathrm{x}\right)\:−\:\mathrm{5}\:\mathrm{f}\:^{'} \left(\mathrm{x}\right)\:+\:\mathrm{6}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{0}\:\:,\:\:\forall\:\mathrm{x}\:\in\:\mathbb{R} \\ $$$$\mathrm{Find}:\:\:\:\boldsymbol{\Omega}\:=\underset{\boldsymbol{\mathrm{x}}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}\:−\:\frac{\mathrm{1}}{\mathrm{f}\:\left(\mathrm{x}\right)}\right)^{\boldsymbol{\mathrm{x}}} \\ $$

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Prove that, { ((X(t) = (1/( (√(1 + t^2 )))) −LN ((1 + (√(1 + t^2 )))/t))),((Y(t) = (t/( (√(1 + t^2 )))))) :} function is the solution of the following equation: y (√(1 + y′^2 )) = y^′

$$\mathrm{Prove}\:\mathrm{that}, \\ $$$$\begin{cases}{\mathrm{X}\left(\mathrm{t}\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:+\:\mathrm{t}^{\mathrm{2}} }}\:−\mathrm{LN}\:\frac{\mathrm{1}\:+\:\sqrt{\mathrm{1}\:+\:\mathrm{t}^{\mathrm{2}} }}{\mathrm{t}}}\\{\mathrm{Y}\left(\mathrm{t}\right)\:=\:\frac{\mathrm{t}}{\:\sqrt{\mathrm{1}\:+\:\mathrm{t}^{\mathrm{2}} }}}\end{cases} \\ $$$$\mathrm{function}\:\mathrm{is}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{equation}: \\ $$$$\mathrm{y}\:\sqrt{\mathrm{1}\:+\:\mathrm{y}'^{\mathrm{2}} }\:=\:\mathrm{y}^{'} \\ $$

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Prove it by mathematical induction: ∣ Σ_(j=1) ^n x_j ∣ ≤ Σ_(j=1) ^n sin x_j ; x_j ∈ [ 0 , π ]

$$\mathrm{Prove}\:\mathrm{it}\:\mathrm{by}\:\mathrm{mathematical}\:\mathrm{induction}: \\ $$$$\mid\:\:\underset{\boldsymbol{\mathrm{j}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\:\mathrm{x}_{\boldsymbol{\mathrm{j}}} \:\:\mid\:\:\leqslant\:\:\underset{\boldsymbol{\mathrm{j}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\:\mathrm{sin}\:\mathrm{x}_{\boldsymbol{\mathrm{j}}} \:\:\:\:\:;\:\:\:\:\:\mathrm{x}_{\boldsymbol{\mathrm{j}}} \:\in\:\left[\:\mathrm{0}\:,\:\pi\:\right] \\ $$

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find the domain and range of y = (1/((x − 1)(x + 2))) restricted to 0 ≤ x ≤ 6

$$\mathrm{find}\:\mathrm{the}\:\mathrm{domain}\:\mathrm{and}\:\mathrm{range}\:\mathrm{of}\:\:\:\mathrm{y}\:\:=\:\:\frac{\mathrm{1}}{\left(\mathrm{x}\:\:−\:\:\mathrm{1}\right)\left(\mathrm{x}\:\:+\:\:\mathrm{2}\right)} \\ $$$$\mathrm{restricted}\:\mathrm{to}\:\:\:\mathrm{0}\:\:\leqslant\:\:\mathrm{x}\:\:\leqslant\:\:\mathrm{6} \\ $$

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