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Question Number 172516    Answers: 0   Comments: 1

solve x^2 =16^x

$${solve} \\ $$$${x}^{\mathrm{2}} =\mathrm{16}^{{x}} \\ $$

Question Number 172515    Answers: 1   Comments: 0

simplify ((z^2 −1)/(z−1))÷(1/(z^2 +1))×(1/(z+(1/z)))

$${simplify} \\ $$$$\frac{{z}^{\mathrm{2}} −\mathrm{1}}{{z}−\mathrm{1}}\boldsymbol{\div}\frac{\mathrm{1}}{{z}^{\mathrm{2}} +\mathrm{1}}×\frac{\mathrm{1}}{{z}+\frac{\mathrm{1}}{{z}}} \\ $$

Question Number 172514    Answers: 0   Comments: 0

if y=bcoslog((x/n))^n then (dy/dx)=??

$${if}\:{y}={bcoslog}\left(\frac{{x}}{{n}}\right)^{{n}} \\ $$$${then}\:\frac{{dy}}{{dx}}=?? \\ $$

Question Number 172513    Answers: 2   Comments: 2

simplify (√(a^2 b+b^3 +2ab^2 )) −(√(a^2 b+4b^3 +4ab^3 ))

$${simplify} \\ $$$$\sqrt{{a}^{\mathrm{2}} {b}+{b}^{\mathrm{3}} +\mathrm{2}{ab}^{\mathrm{2}} }\:−\sqrt{{a}^{\mathrm{2}} {b}+\mathrm{4}{b}^{\mathrm{3}} +\mathrm{4}{ab}^{\mathrm{3}} } \\ $$

Question Number 172511    Answers: 1   Comments: 0

solve: (a+b−2c)x^2 +(2a−b−c)x+(c+a−2b)=0

$${solve}: \\ $$$$\left({a}+{b}−\mathrm{2}{c}\right){x}^{\mathrm{2}} +\left(\mathrm{2}{a}−{b}−{c}\right){x}+\left({c}+{a}−\mathrm{2}{b}\right)=\mathrm{0} \\ $$

Question Number 172510    Answers: 1   Comments: 0

Question Number 172509    Answers: 2   Comments: 2

Question Number 172504    Answers: 2   Comments: 0

Question Number 172498    Answers: 0   Comments: 0

Question Number 172497    Answers: 0   Comments: 0

Question Number 172490    Answers: 0   Comments: 4

Question Number 172488    Answers: 2   Comments: 0

Question Number 172483    Answers: 1   Comments: 0

Question Number 172484    Answers: 1   Comments: 0

Question Number 172480    Answers: 0   Comments: 0

In acute △ABC holds: Π_(cyc) (tan A)^a ≤ Π_(cyc) (cot (A/2))^((b + c)/2)

$$\mathrm{In}\:\mathrm{acute}\:\bigtriangleup\mathrm{ABC}\:\mathrm{holds}: \\ $$$$\underset{\boldsymbol{\mathrm{cyc}}} {\prod}\:\left(\mathrm{tan}\:\mathrm{A}\right)^{\boldsymbol{\mathrm{a}}} \:\:\leqslant\:\:\underset{\boldsymbol{\mathrm{cyc}}} {\prod}\:\left(\mathrm{cot}\:\frac{\mathrm{A}}{\mathrm{2}}\right)^{\frac{\boldsymbol{\mathrm{b}}\:+\:\boldsymbol{\mathrm{c}}}{\mathrm{2}}} \\ $$

Question Number 172468    Answers: 0   Comments: 1

Question Number 172465    Answers: 0   Comments: 1

Question Number 172455    Answers: 1   Comments: 0

using maclaurin′s series exapand (x/2)(((e^x +1)/(e^x −1))) upto x^4 term

$$\mathrm{using}\:\mathrm{maclaurin}'\mathrm{s}\:\mathrm{series}\:\mathrm{exapand} \\ $$$$\frac{{x}}{\mathrm{2}}\left(\frac{{e}^{{x}} +\mathrm{1}}{{e}^{{x}} −\mathrm{1}}\right)\:\mathrm{upto}\:{x}^{\mathrm{4}} \:\mathrm{term} \\ $$

Question Number 172449    Answers: 0   Comments: 0

Question Number 172448    Answers: 1   Comments: 0

Question Number 172453    Answers: 1   Comments: 6

Question Number 172452    Answers: 1   Comments: 0

If , f(x)= ∣ x − k⌊x⌋∣ be one −to−one function on (0,2) find the value of ′′ k ′′.

$$ \\ $$$$\:\:\:\:\:\mathrm{I}{f}\:\:,\:{f}\left({x}\right)=\:\mid\:{x}\:−\:{k}\lfloor{x}\rfloor\mid\:{be}\: \\ $$$$\:{one}\:−{to}−{one}\:{function}\:{on}\:\left(\mathrm{0},\mathrm{2}\right) \\ $$$$\:\:\: \\ $$$$\:\:\:\:{find}\:{the}\:{value}\:{of}\:\:\:\:\:''\:\:\:\:{k}\:\:\:''. \\ $$$$ \\ $$

Question Number 172443    Answers: 2   Comments: 0

Question Number 172439    Answers: 1   Comments: 2

Question Number 172576    Answers: 0   Comments: 0

Question Number 172437    Answers: 1   Comments: 0

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