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Question Number 174677 Answers: 0 Comments: 1

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In △ABC the following relationship holds: a^4 < (b^2 + c^2 )^2 + 9R^4

$$\mathrm{In}\:\:\bigtriangleup\mathrm{ABC}\:\:\mathrm{the}\:\mathrm{following}\:\mathrm{relationship}\:\mathrm{holds}: \\ $$$$\mathrm{a}^{\mathrm{4}} \:<\:\left(\mathrm{b}^{\mathrm{2}} \:+\:\mathrm{c}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\:\mathrm{9R}^{\mathrm{4}} \\ $$

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for what values a and b is the function differentiable at x=2? 4x−2x^(2 ) x≤2 ax^(3 ) +bx,x>2

$${for}\:{what}\:{values}\:{a}\:{and}\:{b}\:{is}\:{the}\: \\ $$$${function}\:{differentiable}\:{at}\:{x}=\mathrm{2}? \\ $$$$\mathrm{4}{x}−\mathrm{2}{x}^{\mathrm{2}\:} {x}\leqslant\mathrm{2} \\ $$$${ax}^{\mathrm{3}\:} +{bx},{x}>\mathrm{2} \\ $$

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find the domain and range of tbe following 1) y = cot(2x+1) 2) y = sin∣x∣ 3) ∣y∣ = cosx 4) y = cos(cosx) 5) y =∣cscx∣ −1 6) y = sin(sinx)

$${find}\:{the}\:{domain}\:{and}\:{range}\:{of}\:{tbe}\:{following} \\ $$$$ \\ $$$$\left.\mathrm{1}\right)\:{y}\:=\:{cot}\left(\mathrm{2}{x}+\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right)\:{y}\:=\:{sin}\mid{x}\mid \\ $$$$\left.\mathrm{3}\right)\:\mid{y}\mid\:=\:{cosx} \\ $$$$\left.\mathrm{4}\right)\:{y}\:=\:{cos}\left({cosx}\right) \\ $$$$\left.\mathrm{5}\right)\:{y}\:=\mid{cscx}\mid\:−\mathrm{1} \\ $$$$\left.\mathrm{6}\right)\:{y}\:=\:{sin}\left({sinx}\right) \\ $$

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After being marked down 20 percent. a calculator sells for $10. The Original selling price was ?

$$\mathrm{After}\:\mathrm{being}\:\mathrm{marked}\:\mathrm{down}\:\mathrm{20}\:\mathrm{percent}. \\ $$$$\mathrm{a}\:\mathrm{calculator}\:\mathrm{sells}\:\mathrm{for}\:\$\mathrm{10}.\:\mathrm{The}\:\mathrm{Original} \\ $$$$\mathrm{selling}\:\mathrm{price}\:\mathrm{was}\:? \\ $$

Question Number 174655 Answers: 1 Comments: 0

Factorize a^3 − 16a − 3

$$\mathrm{Factorize}\:{a}^{\mathrm{3}} \:−\:\mathrm{16}{a}\:−\:\mathrm{3} \\ $$

Question Number 174654 Answers: 0 Comments: 1

Factorize x^9 + x^7 + 1

$$\mathrm{Factorize}\:{x}^{\mathrm{9}} \:+\:{x}^{\mathrm{7}} \:+\:\mathrm{1} \\ $$

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Ω = ∫ (x/(1+csc x)) dx =?

$$\:\:\:\Omega\:=\:\int\:\frac{{x}}{\mathrm{1}+\mathrm{csc}\:{x}}\:{dx}\:=? \\ $$

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In a mixture of Skettles and M&M′s, 80% of the pieces are M&M′s. A fourth of this mixture is replaced by a second mixture, resulting in combination which contain 16% Skittles in total. What was the percentage of Skittles in the second mixture?

$$\mathrm{In}\:\mathrm{a}\:\mathrm{mixture}\:\mathrm{of}\:\:\mathrm{Skettles}\:\mathrm{and}\:\mathrm{M\&M}'\mathrm{s}, \\ $$$$\mathrm{80\%}\:\mathrm{of}\:\mathrm{the}\:\mathrm{pieces}\:\mathrm{are}\:\mathrm{M\&M}'\mathrm{s}.\:\mathrm{A}\:\mathrm{fourth} \\ $$$$\mathrm{of}\:\mathrm{this}\:\mathrm{mixture}\:\mathrm{is}\:\mathrm{replaced}\:\mathrm{by}\:\mathrm{a}\:\mathrm{second} \\ $$$$\mathrm{mixture},\:\mathrm{resulting}\:\mathrm{in}\:\mathrm{combination} \\ $$$$\mathrm{which}\:\mathrm{contain}\:\mathrm{16\%}\:\mathrm{Skittles}\:\mathrm{in}\:\mathrm{total}. \\ $$$$\mathrm{What}\:\mathrm{was}\:\mathrm{the}\:\mathrm{percentage}\:\mathrm{of}\:\mathrm{Skittles} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{second}\:\mathrm{mixture}? \\ $$

Question Number 174612 Answers: 0 Comments: 1

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Find the expected payback for a game in which you bet $8.00 on any number from 0 to 99 and if your number comes up, you win $2,000.00

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{expected}\:\mathrm{payback}\:\mathrm{for}\:\mathrm{a}\: \\ $$$$\mathrm{game}\:\mathrm{in}\:\mathrm{which}\:\mathrm{you}\:\mathrm{bet}\:\$\mathrm{8}.\mathrm{00}\:\mathrm{on}\:\mathrm{any} \\ $$$$\mathrm{number}\:\mathrm{from}\:\mathrm{0}\:\mathrm{to}\:\mathrm{99}\:\mathrm{and}\:\mathrm{if}\:\mathrm{your} \\ $$$$\mathrm{number}\:\mathrm{comes}\:\mathrm{up},\:\mathrm{you}\:\mathrm{win}\:\$\mathrm{2},\mathrm{000}.\mathrm{00} \\ $$

Question Number 174610 Answers: 0 Comments: 0

find the value of this integral: I=∫_0 ^∞ ((tan^(−1) (x))/(1+x+x^2 )) dx

$${find}\:{the}\:{value}\:{of}\:{this}\:{integral}: \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \:\frac{{tan}^{−\mathrm{1}} \left({x}\right)}{\mathrm{1}+{x}+{x}^{\mathrm{2}} }\:{dx} \\ $$

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