Question and Answers Forum

All Questions   Topic List

AlgebraQuestion and Answers: Page 5

Question Number 224029    Answers: 2   Comments: 0

Question Number 224028    Answers: 1   Comments: 0

Question Number 224018    Answers: 2   Comments: 0

x ≠ y λ ≥ 1 { ((x + λ^2 = (y − λ)^2 )),((y + λ^2 = (x − λ)^2 )) :} Find: (((x^2 + y^2 )/(4λ^2 − 1)))^(2025) = ?

$$\mathrm{x}\:\neq\:\mathrm{y} \\ $$$$\lambda\:\geqslant\:\mathrm{1} \\ $$$$\begin{cases}{\mathrm{x}\:+\:\lambda^{\mathrm{2}} \:=\:\left(\mathrm{y}\:−\:\lambda\right)^{\mathrm{2}} }\\{\mathrm{y}\:+\:\lambda^{\mathrm{2}} \:=\:\left(\mathrm{x}\:−\:\lambda\right)^{\mathrm{2}} }\end{cases} \\ $$$$\mathrm{Find}:\:\:\:\left(\frac{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} }{\mathrm{4}\lambda^{\mathrm{2}} \:−\:\mathrm{1}}\right)^{\mathrm{2025}} =\:\:? \\ $$

Question Number 224017    Answers: 0   Comments: 0

x,y,z>0 xy+yz+zx+2xyz=1 prove that: (√(1−x^2 )) + (√(1−y^2 )) + (√(1−z^2 )) ≤ ((3 (√3))/2)

$$\mathrm{x},\mathrm{y},\mathrm{z}>\mathrm{0} \\ $$$$\mathrm{xy}+\mathrm{yz}+\mathrm{zx}+\mathrm{2xyz}=\mathrm{1} \\ $$$$\mathrm{prove}\:\mathrm{that}: \\ $$$$\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:+\:\sqrt{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }\:+\:\sqrt{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }\:\leqslant\:\frac{\mathrm{3}\:\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$

Question Number 224016    Answers: 0   Comments: 0

a,b,c>0 a+b+c+2=abc prove that: (1/( (√(7+a)))) + (1/( (√(7+b)))) + (1/( (√(7+c)))) ≤ 1

$$\mathrm{a},\mathrm{b},\mathrm{c}>\mathrm{0} \\ $$$$\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{2}=\mathrm{abc} \\ $$$$\mathrm{prove}\:\mathrm{that}:\:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}+\mathrm{a}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}+\mathrm{b}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}+\mathrm{c}}}\:\leqslant\:\mathrm{1} \\ $$

Question Number 224015    Answers: 0   Comments: 0

a,b,c>0 a+b+c+2=abc prove that: (√a) + (√b) + (√c) ≤ (3/2) (√(abc))

$$\mathrm{a},\mathrm{b},\mathrm{c}>\mathrm{0} \\ $$$$\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{2}=\mathrm{abc} \\ $$$$\mathrm{prove}\:\mathrm{that}:\:\:\:\sqrt{\mathrm{a}}\:+\:\sqrt{\mathrm{b}}\:+\:\sqrt{\mathrm{c}}\:\leqslant\:\frac{\mathrm{3}}{\mathrm{2}}\:\sqrt{\mathrm{abc}} \\ $$

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

$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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: 3   Comments: 0

Question Number 223508    Answers: 0   Comments: 0

  Pg 1      Pg 2      Pg 3      Pg 4      Pg 5      Pg 6      Pg 7      Pg 8      Pg 9      Pg 10   

Terms of Service

Privacy Policy

Contact: info@tinkutara.com