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AlgebraQuestion and Answers: Page 2

Question Number 223995    Answers: 1   Comments: 0

Question Number 223965    Answers: 2   Comments: 0

Question Number 223964    Answers: 3   Comments: 0

Question Number 223858    Answers: 1   Comments: 0

Question Number 223823    Answers: 3   Comments: 0

(√(4x+1))+(√(3x−2))=1 x=?

$$\sqrt{\mathrm{4}{x}+\mathrm{1}}+\sqrt{\mathrm{3}{x}−\mathrm{2}}=\mathrm{1} \\ $$$${x}=? \\ $$

Question Number 223822    Answers: 2   Comments: 0

(((4/3))^(4/3) ) Rewrite in simplest radical form

$$\left(\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} \right) \\ $$$$\:{Rewrite}\:{in}\:{simplest}\:{radical}\:{form} \\ $$

Question Number 223804    Answers: 0   Comments: 3

Question Number 223800    Answers: 1   Comments: 0

Given f(x)= ((x^2 +14x+40)/(g(x)))−43 h(x)= ((g(x)+51)/(x+4)) m(x)= ((h(x)−9)/(x−2)) , x≠2 m(2)= 2043. If f(x) divided by x^2 +8x−20 gives remainder is M(x)=ax+b then the value of M(98)=?

$$\:\mathrm{Given}\:\mathrm{f}\left(\mathrm{x}\right)=\:\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{14x}+\mathrm{40}}{\mathrm{g}\left(\mathrm{x}\right)}−\mathrm{43} \\ $$$$\:\mathrm{h}\left(\mathrm{x}\right)=\:\frac{\mathrm{g}\left(\mathrm{x}\right)+\mathrm{51}}{\mathrm{x}+\mathrm{4}} \\ $$$$\:\mathrm{m}\left(\mathrm{x}\right)=\:\frac{\mathrm{h}\left(\mathrm{x}\right)−\mathrm{9}}{\mathrm{x}−\mathrm{2}}\:,\:\mathrm{x}\neq\mathrm{2} \\ $$$$\:\mathrm{m}\left(\mathrm{2}\right)=\:\mathrm{2043}.\: \\ $$$$\:\mathrm{If}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{divided}\:\mathrm{by}\:\mathrm{x}^{\mathrm{2}} +\mathrm{8x}−\mathrm{20}\: \\ $$$$\:\mathrm{gives}\:\mathrm{remainder}\:\mathrm{is}\:\mathrm{M}\left(\mathrm{x}\right)=\mathrm{ax}+\mathrm{b} \\ $$$$\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{M}\left(\mathrm{98}\right)=?\: \\ $$

Question Number 223734    Answers: 1   Comments: 0

Question Number 223703    Answers: 2   Comments: 0

Question Number 223700    Answers: 1   Comments: 0

$$\:\underline{\underbrace{\:}} \\ $$

Question Number 223685    Answers: 1   Comments: 0

25^x −8.5^x =−16

$$\mathrm{25}^{{x}} −\mathrm{8}.\mathrm{5}^{{x}} =−\mathrm{16} \\ $$

Question Number 223655    Answers: 3   Comments: 4

Question Number 223636    Answers: 0   Comments: 0

Factor the following expression: (((arctan(x^5 +1)))^(1/5) )^x^(−x^2 )

$$\mathrm{Factor}\:\mathrm{the}\:\mathrm{following}\:\mathrm{expression}: \\ $$$$\left(\sqrt[{\mathrm{5}}]{\mathrm{arctan}\left({x}^{\mathrm{5}} +\mathrm{1}\right)}\right)^{{x}^{−{x}^{\mathrm{2}} } } \\ $$

Question Number 223626    Answers: 1   Comments: 6

{ ((x^2 +y^2 +xy=25)),((y^2 +z^2 +yz=49)),((z^2 +x^2 +zx=64)) :} (x+y+z)^2 −100=??

$$\begin{cases}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}=\mathrm{25}}\\{{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +{yz}=\mathrm{49}}\\{{z}^{\mathrm{2}} +{x}^{\mathrm{2}} +{zx}=\mathrm{64}}\end{cases} \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{2}} −\mathrm{100}=?? \\ $$

Question Number 223615    Answers: 3   Comments: 0

40^(x−1) =2^(2x+1)

$$\mathrm{40}^{{x}−\mathrm{1}} =\mathrm{2}^{\mathrm{2}{x}+\mathrm{1}} \\ $$

Question Number 223585    Answers: 1   Comments: 1

Question Number 223571    Answers: 2   Comments: 0

S_1 = 1∙1! + 2∙2! + 3∙3! +...+ 16∙16! S_2 = 1∙1! + 2∙2! + 3∙3! +...+ 14∙14! Find: (S_1 /S_2 ) = ?

$$\mathrm{S}_{\mathrm{1}} \:=\:\mathrm{1}\centerdot\mathrm{1}!\:+\:\mathrm{2}\centerdot\mathrm{2}!\:+\:\mathrm{3}\centerdot\mathrm{3}!\:+...+\:\mathrm{16}\centerdot\mathrm{16}! \\ $$$$\mathrm{S}_{\mathrm{2}} \:=\:\mathrm{1}\centerdot\mathrm{1}!\:+\:\mathrm{2}\centerdot\mathrm{2}!\:+\:\mathrm{3}\centerdot\mathrm{3}!\:+...+\:\mathrm{14}\centerdot\mathrm{14}! \\ $$$$\mathrm{Find}:\:\:\:\frac{\mathrm{S}_{\mathrm{1}} }{\mathrm{S}_{\mathrm{2}} }\:=\:? \\ $$

Question Number 223569    Answers: 2   Comments: 0

Question Number 223508    Answers: 0   Comments: 0

Question Number 223487    Answers: 1   Comments: 0

Let Φ be the hyperbola xy = b², b ≠ 0, and P be a point on Φ. Let Q be the image of reflection of P about the origin. Construct a circle ω centred at P with radius PQ. ω cuts Φ at the points B, C, D, Q. Prove that ΔBCD is equilateral, no matter what the value of b is.

Let Φ be the hyperbola xy = b², b ≠ 0, and P be a point on Φ. Let Q be the image of reflection of P about the origin. Construct a circle ω centred at P with radius PQ. ω cuts Φ at the points B, C, D, Q. Prove that ΔBCD is equilateral, no matter what the value of b is.

Question Number 223483    Answers: 1   Comments: 0

(x−1)(x−2)=1 (x−1)^(15) −(1/((x−1)^(15) ))=??

$$\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)=\mathrm{1} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{15}} −\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{15}} }=?? \\ $$

Question Number 223474    Answers: 3   Comments: 1

(√(1+(√(1+x))))=(x)^(1/3)

$$\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+{x}}}=\sqrt[{\mathrm{3}}]{{x}} \\ $$

Question Number 223459    Answers: 4   Comments: 0

solve for x∈R (√(25−10x−x^2 ))+(√(15−x^2 ))=2(√5)

$${solve}\:{for}\:{x}\in{R} \\ $$$$\sqrt{\mathrm{25}−\mathrm{10}{x}−{x}^{\mathrm{2}} }+\sqrt{\mathrm{15}−{x}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{5}} \\ $$

Question Number 223429    Answers: 2   Comments: 1

....

$$.... \\ $$

Question Number 223424    Answers: 1   Comments: 3

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