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AlgebraQuestion and Answers: Page 4

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: 3 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

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Question Number 223424 Answers: 1 Comments: 3

Question Number 223386 Answers: 1 Comments: 0

Question Number 223374 Answers: 3 Comments: 0

Question Number 223405 Answers: 2 Comments: 0

Question Number 223403 Answers: 1 Comments: 1

Question Number 223400 Answers: 1 Comments: 0

f(1)=2025 𝚺_1 ^n f(k)=n^2 .f(n) f(2025)=?

$$\boldsymbol{{f}}\left(\mathrm{1}\right)=\mathrm{2025} \\ $$$$\underset{\mathrm{1}} {\overset{\boldsymbol{{n}}} {\boldsymbol{\sum}}{f}}\left(\boldsymbol{{k}}\right)=\boldsymbol{{n}}^{\mathrm{2}} .\boldsymbol{{f}}\left(\boldsymbol{{n}}\right) \\ $$$$\boldsymbol{{f}}\left(\mathrm{2025}\right)=? \\ $$

Question Number 223354 Answers: 0 Comments: 6

Question Number 223267 Answers: 3 Comments: 0

r,s,t ; are the roots of: x^3 +5x+1=0 find: (r^3 −1)(s^3 −1)(t^3 −1)

$$\boldsymbol{{r}},\boldsymbol{{s}},\boldsymbol{{t}}\:;\:\boldsymbol{{are}}\:\boldsymbol{{the}}\:\boldsymbol{{roots}}\:\boldsymbol{{of}}: \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{{x}}^{\mathrm{3}} +\mathrm{5}\boldsymbol{{x}}+\mathrm{1}=\mathrm{0} \\ $$$$\boldsymbol{{find}}: \\ $$$$\:\:\:\:\:\:\left(\boldsymbol{{r}}^{\mathrm{3}} −\mathrm{1}\right)\left(\boldsymbol{{s}}^{\mathrm{3}} −\mathrm{1}\right)\left(\boldsymbol{{t}}^{\mathrm{3}} −\mathrm{1}\right) \\ $$

Question Number 223261 Answers: 0 Comments: 1

Question Number 223240 Answers: 1 Comments: 0

x^x =i number of solutions??

$${x}^{{x}} ={i} \\ $$$${number}\:{of}\:{solutions}?? \\ $$

Question Number 223230 Answers: 3 Comments: 0

Maksimum (((x^2 − 4x + 1)/(x^2 + 1))) = ?

$$\mathrm{Maksimum}\:\left(\frac{\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{4x}\:+\:\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1}}\right)\:=\:? \\ $$

Question Number 223216 Answers: 2 Comments: 0

Question Number 223203 Answers: 2 Comments: 0

Find x and y ix+y=ix^3 −y^3 and xy(y−ix)=c

$${Find}\:{x}\:{and}\:{y} \\ $$$$\:\:\:\:{ix}+{y}={ix}^{\mathrm{3}} −{y}^{\mathrm{3}} \\ $$$$\:\:\:{and}\: \\ $$$$\:\:\:\:{xy}\left({y}−{ix}\right)={c} \\ $$

Question Number 223189 Answers: 1 Comments: 3

(x−1)(x^2 −2)(x^3 −3)(x^4 −4)=36 x=?

$$\left(\boldsymbol{{x}}−\mathrm{1}\right)\left(\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{2}\right)\left(\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{3}\right)\left(\boldsymbol{{x}}^{\mathrm{4}} −\mathrm{4}\right)=\mathrm{36} \\ $$$$\boldsymbol{{x}}=? \\ $$

Question Number 223179 Answers: 1 Comments: 1

If: xy = e^(𝛑/4) Find: tg (ln ((x^3 /y))) ∙ tg (ln ((y^3 /x))) = ?

$$\mathrm{If}:\:\:\:\mathrm{xy}\:=\:\boldsymbol{\mathrm{e}}^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{tg}\:\left(\mathrm{ln}\:\left(\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{y}}\right)\right)\:\centerdot\:\mathrm{tg}\:\left(\mathrm{ln}\:\left(\frac{\mathrm{y}^{\mathrm{3}} }{\mathrm{x}}\right)\right)\:=\:? \\ $$

Question Number 223161 Answers: 2 Comments: 4

Solve for x if (√(x+1))+(√(x−1))=1

$${Solve}\:{for}\:{x}\:{if} \\ $$$$\:\sqrt{{x}+\mathrm{1}}+\sqrt{{x}−\mathrm{1}}=\mathrm{1} \\ $$

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