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AllQuestion and Answers: Page 529

Question Number 165068    Answers: 0   Comments: 1

Question Number 165063    Answers: 1   Comments: 1

lim_(x→0) ((x^(tanx) −cosx)/(x^(sinx) −e^x ))

$$\:{lim}_{{x}\rightarrow\mathrm{0}} \frac{{x}^{{tanx}} −{cosx}}{{x}^{{sinx}} −{e}^{{x}} } \\ $$

Question Number 165057    Answers: 1   Comments: 1

Question Number 165054    Answers: 1   Comments: 0

Question Number 165060    Answers: 1   Comments: 0

x−(1/x)=4 then x^2 −(1/x^2 )=?

$${x}−\frac{\mathrm{1}}{{x}}=\mathrm{4}\:\:{then}\:{x}^{\mathrm{2}} −\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=? \\ $$

Question Number 165044    Answers: 0   Comments: 2

If n = j + k, prove or disprove (d^n /dx^n )f(x) = (d^j /dx^j )((d^k /dx^k )f(x)) = (d^k /dx^k )((d^j /dx^j )f(x))

$$\mathrm{If}\:{n}\:=\:{j}\:+\:{k},\:\mathrm{prove}\:\mathrm{or}\:\mathrm{disprove} \\ $$$$\frac{{d}^{{n}} }{{dx}^{{n}} }{f}\left({x}\right)\:=\:\frac{{d}^{{j}} }{{dx}^{{j}} }\left(\frac{{d}^{{k}} }{{dx}^{{k}} }{f}\left({x}\right)\right)\:=\:\frac{{d}^{{k}} }{{dx}^{{k}} }\left(\frac{{d}^{{j}} }{{dx}^{{j}} }{f}\left({x}\right)\right) \\ $$

Question Number 165010    Answers: 0   Comments: 0

Ω= Σ_(n=1) ^∞ (( ψ^((2)) ( 1+ n ))/( n)) = ? −−− m.n −−−

$$ \\ $$$$\:\:\:\:\:\:\:\Omega=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\psi\:^{\left(\mathrm{2}\right)} \left(\:\mathrm{1}+\:{n}\:\right)}{\:{n}}\:=\:? \\ $$$$\:\:\:\:\:\:−−−\:{m}.{n}\:−−− \\ $$$$\:\:\:\: \\ $$

Question Number 165006    Answers: 0   Comments: 1

Question Number 165024    Answers: 1   Comments: 7

Question Number 165021    Answers: 2   Comments: 0

fog=6x^2 −2x+3 gof=12x^2 −14x+6 f(2)=?

$${fog}=\mathrm{6}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3} \\ $$$${gof}=\mathrm{12}{x}^{\mathrm{2}} −\mathrm{14}{x}+\mathrm{6} \\ $$$${f}\left(\mathrm{2}\right)=? \\ $$

Question Number 165019    Answers: 0   Comments: 0

Question Number 165018    Answers: 0   Comments: 0

y = Γ(m+n) Find (dy/dn)

$${y}\:=\:\Gamma\left({m}+{n}\right)\: \\ $$$${Find}\:\frac{{dy}}{{dn}} \\ $$

Question Number 165017    Answers: 0   Comments: 0

Question Number 165015    Answers: 2   Comments: 2

Question Number 164996    Answers: 1   Comments: 0

Question Number 164995    Answers: 1   Comments: 0

Question Number 164993    Answers: 1   Comments: 0

solve by series ∫_0 ^( ∞) ((sinx)/x) dx

$${solve}\:{by}\:{series}\:\int_{\mathrm{0}} ^{\:\infty} \:\frac{{sinx}}{{x}}\:{dx} \\ $$

Question Number 164992    Answers: 0   Comments: 0

Question Number 164991    Answers: 2   Comments: 1

Solve by resideo theorem ∫_(−∞) ^( ∞) (z^2 /(z^4 +1)) dz

$$\boldsymbol{{Solve}}\:\boldsymbol{{by}}\:\boldsymbol{{resideo}}\:\boldsymbol{{theorem}}\:\int_{−\infty} ^{\:\infty} \:\frac{\boldsymbol{{z}}^{\mathrm{2}} }{\boldsymbol{{z}}^{\mathrm{4}} +\mathrm{1}}\:\boldsymbol{{dz}} \\ $$

Question Number 164985    Answers: 2   Comments: 0

Question Number 164982    Answers: 0   Comments: 0

Prove, that (a/b) : (c/d) = (a/b) ∙ (d/c)

$$\mathrm{Prove},\:\mathrm{that}\:\frac{{a}}{{b}}\::\:\frac{{c}}{{d}}\:=\:\frac{{a}}{{b}}\:\centerdot\:\frac{{d}}{{c}} \\ $$

Question Number 164984    Answers: 0   Comments: 0

Question Number 164974    Answers: 1   Comments: 0

∫_0 ^( 1) (( ln( 1− x ).ln(x ) )/x^( (3/2)) )dx=^? π^( 2) −8ln(2 ) −−− m.n −−−

$$ \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\mathrm{ln}\left(\:\mathrm{1}−\:{x}\:\right).\mathrm{ln}\left({x}\:\right)\:\:}{{x}^{\:\frac{\mathrm{3}}{\mathrm{2}}} }{dx}\overset{?} {=}\pi^{\:\mathrm{2}} −\mathrm{8ln}\left(\mathrm{2}\:\right)\: \\ $$$$\:\:\:\:\:−−−\:\:{m}.{n}\:−−− \\ $$$$ \\ $$

Question Number 164973    Answers: 1   Comments: 0

Question Number 164970    Answers: 0   Comments: 0

⌊x(x^(2+1) /(x/(10−3^x ))) + x(x^(2+2) /(x/(9−3^x ))) + x(x^(2+3) /(x/(8−3^x ))) +x (x^(2+4) /(x/(7−3^x ))) + x(x^(2+5) /(x/(6−3^x ))) ≥ (1/((25)/x^x^(x ((1/(25)))) ))⌋ {Z.A}

$$\:\:\lfloor\boldsymbol{{x}}\frac{\boldsymbol{{x}}^{\mathrm{2}+\mathrm{1}} }{\frac{\boldsymbol{{x}}}{\mathrm{10}−\mathrm{3}^{\boldsymbol{{x}}} }}\:+\:\boldsymbol{{x}}\frac{\boldsymbol{{x}}^{\mathrm{2}+\mathrm{2}} }{\frac{\boldsymbol{{x}}}{\mathrm{9}−\mathrm{3}^{\boldsymbol{{x}}} }}\:+\:\boldsymbol{{x}}\frac{\boldsymbol{{x}}^{\mathrm{2}+\mathrm{3}} }{\frac{\boldsymbol{{x}}}{\mathrm{8}−\mathrm{3}^{\boldsymbol{{x}}} }}\:+\boldsymbol{{x}}\:\frac{\boldsymbol{{x}}^{\mathrm{2}+\mathrm{4}} }{\frac{\boldsymbol{{x}}}{\mathrm{7}−\mathrm{3}^{\boldsymbol{{x}}} }}\:+\:\boldsymbol{{x}}\frac{\boldsymbol{{x}}^{\mathrm{2}+\mathrm{5}} }{\frac{\boldsymbol{{x}}}{\mathrm{6}−\mathrm{3}^{\boldsymbol{{x}}} }}\:\geqslant\:\frac{\mathrm{1}}{\frac{\mathrm{25}}{\boldsymbol{{x}}^{\boldsymbol{{x}}^{\boldsymbol{{x}}\:\left(\frac{\mathrm{1}}{\mathrm{25}}\right)} } }}\rfloor \\ $$$$\:\:\:\left\{\mathrm{Z}.\mathrm{A}\right\} \\ $$$$\: \\ $$

Question Number 164966    Answers: 1   Comments: 0

⌊((5/x) + (4/x) + (3/x) + (2/x) + (1/x)) • ((x/1) − (x/2) − (x/3) − (x/4) − (x/5) )^2 > (1/(15))⌋ {za}

$$\:\:\:\:\:\lfloor\left(\frac{\mathrm{5}}{\boldsymbol{{x}}}\:+\:\frac{\mathrm{4}}{\boldsymbol{{x}}}\:+\:\frac{\mathrm{3}}{\boldsymbol{{x}}}\:+\:\frac{\mathrm{2}}{\boldsymbol{{x}}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{x}}}\right)\:\bullet\:\left(\frac{\boldsymbol{{x}}}{\mathrm{1}}\:−\:\frac{\boldsymbol{{x}}}{\mathrm{2}}\:−\:\frac{\boldsymbol{{x}}}{\mathrm{3}}\:\:−\:\frac{\boldsymbol{{x}}}{\mathrm{4}}\:−\:\frac{\boldsymbol{{x}}}{\mathrm{5}}\:\right)^{\mathrm{2}} >\:\frac{\mathrm{1}}{\mathrm{15}}\rfloor \\ $$$$\:\:\:\left\{\mathrm{za}\right\} \\ $$

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