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Question Number 193377    Answers: 1   Comments: 1

Evaluate I=∫_0 ^( ∞) (1/(x^5 +x^4 +x^3 +x^2 +x+1))dx

$${Evaluate} \\ $$$${I}=\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}}{{x}^{\mathrm{5}} +{x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx} \\ $$

Question Number 193375    Answers: 0   Comments: 0

Quantitative Reasoning Example: P_4 +M_3 =16 M_4 −P_3 =18, P_5 +M_3 =2 Find: P_6 +M_4 =? P_5 −M_8 =? M_3 +M_2 =? M_5 −P_(10) =?

$${Quantitative}\:{Reasoning} \\ $$$${Example}:\:\:{P}_{\mathrm{4}} +{M}_{\mathrm{3}} =\mathrm{16} \\ $$$${M}_{\mathrm{4}} −{P}_{\mathrm{3}} =\mathrm{18},\:{P}_{\mathrm{5}} +{M}_{\mathrm{3}} =\mathrm{2} \\ $$$${Find}: \\ $$$${P}_{\mathrm{6}} +{M}_{\mathrm{4}} =? \\ $$$${P}_{\mathrm{5}} −{M}_{\mathrm{8}} =? \\ $$$${M}_{\mathrm{3}} +{M}_{\mathrm{2}} =? \\ $$$${M}_{\mathrm{5}} −{P}_{\mathrm{10}} =? \\ $$$$ \\ $$

Question Number 193371    Answers: 3   Comments: 0

Reduce to first order and solve , showing each step in detail. 1. y′′ +(y′)^3 siny=0 2. y′′=1+(y′)^2

$$\mathrm{Reduce}\:\mathrm{to}\:\mathrm{first}\:\mathrm{order}\:\mathrm{and}\:\mathrm{solve}\:, \\ $$$$\mathrm{showing}\:\mathrm{each}\:\mathrm{step}\:\mathrm{in}\:\mathrm{detail}. \\ $$$$\mathrm{1}.\:\mathrm{y}''\:+\left(\mathrm{y}'\right)^{\mathrm{3}} \mathrm{siny}=\mathrm{0} \\ $$$$\mathrm{2}.\:\mathrm{y}''=\mathrm{1}+\left(\mathrm{y}'\right)^{\mathrm{2}} \\ $$$$ \\ $$

Question Number 193368    Answers: 2   Comments: 0

If log_a y = (1/3) and log_8 a = x + 1 then show that y = 2^(x + 1)

$$\mathrm{If}\:\mathrm{log}_{{a}} {y}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{and}\:\mathrm{log}_{\mathrm{8}} {a}\:=\:{x}\:+\:\mathrm{1}\:\mathrm{then}\:\mathrm{show} \\ $$$$\mathrm{that}\:{y}\:=\:\mathrm{2}^{{x}\:+\:\mathrm{1}} \\ $$

Question Number 193366    Answers: 1   Comments: 0

2sin^2 2x>3cos x+3

$$\mathrm{2sin}\:^{\mathrm{2}} \mathrm{2}{x}>\mathrm{3cos}\:{x}+\mathrm{3} \\ $$

Question Number 193363    Answers: 1   Comments: 0

y = 4 × 10^(2x) Express x in terms of y, giving an exact simplified answer in terms of log base 10.

$${y}\:=\:\mathrm{4}\:×\:\mathrm{10}^{\mathrm{2}{x}} \\ $$$$\mathrm{Express}\:{x}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{y},\:\mathrm{giving}\:\mathrm{an}\:\mathrm{exact} \\ $$$$\mathrm{simplified}\:\mathrm{answer}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{log}\:\mathrm{base}\:\mathrm{10}. \\ $$

Question Number 193360    Answers: 2   Comments: 0

If x = 2^p and y = 4^q then prove that log_2 (x^3 y) = 3p + 2q

$$\mathrm{If}\:{x}\:=\:\mathrm{2}^{{p}} \:\mathrm{and}\:{y}\:=\:\mathrm{4}^{{q}} \:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{log}_{\mathrm{2}} \left({x}^{\mathrm{3}} {y}\right)\:=\:\mathrm{3}{p}\:+\:\mathrm{2}{q} \\ $$

Question Number 193356    Answers: 2   Comments: 0

Question Number 193351    Answers: 0   Comments: 0

Question Number 193350    Answers: 0   Comments: 0

Question Number 193349    Answers: 1   Comments: 0

Question Number 193346    Answers: 2   Comments: 0

(√(4x−3))−(√(2x−5))=2 solve for x

$$\sqrt{\mathrm{4x}−\mathrm{3}}−\sqrt{\mathrm{2x}−\mathrm{5}}=\mathrm{2} \\ $$$$\mathrm{solve}\:\mathrm{for}\:\mathrm{x} \\ $$

Question Number 193345    Answers: 1   Comments: 0

Question Number 193336    Answers: 1   Comments: 1

Question Number 193333    Answers: 0   Comments: 1

Question Number 193332    Answers: 0   Comments: 0

Question Number 193331    Answers: 0   Comments: 0

Question Number 193328    Answers: 1   Comments: 0

f(x)= { ((((x^2 −x)/(x^2 −1)) ; x≠1)),((2x+1; x=1)) :} thene find lim_(x→1) f(x)=?

$$\mathrm{f}\left(\mathrm{x}\right)=\begin{cases}{\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{x}}{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\:;\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}\neq\mathrm{1}}\\{\mathrm{2x}+\mathrm{1};\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}=\mathrm{1}}\end{cases} \\ $$$$\mathrm{thene}\:\mathrm{find}\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}{f}\left({x}\right)=? \\ $$

Question Number 193329    Answers: 0   Comments: 0

Question Number 193323    Answers: 1   Comments: 0

Question Number 193339    Answers: 1   Comments: 0

Prove that a group G of prime order is cyclic.

$${Prove}\:{that}\:{a}\:{group}\:{G}\:{of}\:{prime}\:{order}\:{is}\:{cyclic}. \\ $$$$ \\ $$

Question Number 193316    Answers: 2   Comments: 1

IS THIS RIGHT? I=∫_0 ^∞ e^(−ix^2 ) dx I^2 =∫_0 ^∞ e^(−ix^2 ) dx∫_0 ^∞ e^(−iy^2 ) dy =∫_0 ^∞ ∫_0 ^∞ e^(−iy^2 ) dye^(−ix^2 ) dx =∫_0 ^∞ ∫_0 ^∞ e^(−i(x^2 +y^2 )) dydx dydx=dA=rdrdθ I^2 =∫_0 ^(π/2) ∫_0 ^∞ e^(−ir^2 ) rdrdθ ∫_0 ^∞ e^(−ir^2 ) rdr; u=−ir^2 ⇒du=−2irdr −(i/2)∫_(−∞) ^0 e^u du=−(i/2) I^2 =∫_0 ^(π/2) −(i/2)dθ=−((iπ)/4)⇒I=(√((−iπ)/4)) I=(i/2)(√(e^(iπ/2) π))=((ie^(iπ/4) )/2)(√π) I=((i(√π))/2)(((√2)/2)(1+i))=((i(√(2π)))/4)(1+i)

$$ \\ $$$$\mathrm{IS}\:\mathrm{THIS}\:\mathrm{RIGHT}? \\ $$$$\mathrm{I}=\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{ix}^{\mathrm{2}} } \mathrm{dx} \\ $$$$\mathrm{I}^{\mathrm{2}} =\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{ix}^{\mathrm{2}} } \mathrm{dx}\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{iy}^{\mathrm{2}} } \mathrm{dy} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{iy}^{\mathrm{2}} } \mathrm{dye}^{−\mathrm{ix}^{\mathrm{2}} } \mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{i}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)} \mathrm{dydx} \\ $$$$\mathrm{dydx}=\mathrm{dA}=\mathrm{rdrd}\theta \\ $$$$\mathrm{I}^{\mathrm{2}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{ir}^{\mathrm{2}} } \mathrm{rdrd}\theta \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{ir}^{\mathrm{2}} } \mathrm{rdr};\:\mathrm{u}=−\mathrm{ir}^{\mathrm{2}} \Rightarrow\mathrm{du}=−\mathrm{2irdr} \\ $$$$−\frac{\mathrm{i}}{\mathrm{2}}\int_{−\infty} ^{\mathrm{0}} \mathrm{e}^{\mathrm{u}} \mathrm{du}=−\frac{\mathrm{i}}{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{I}^{\mathrm{2}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\frac{\mathrm{i}}{\mathrm{2}}\mathrm{d}\theta=−\frac{\mathrm{i}\pi}{\mathrm{4}}\Rightarrow\mathrm{I}=\sqrt{\frac{−\mathrm{i}\pi}{\mathrm{4}}} \\ $$$$\mathrm{I}=\frac{\mathrm{i}}{\mathrm{2}}\sqrt{\mathrm{e}^{\mathrm{i}\pi/\mathrm{2}} \pi}=\frac{\mathrm{ie}^{\mathrm{i}\pi/\mathrm{4}} }{\mathrm{2}}\sqrt{\pi} \\ $$$$\mathrm{I}=\frac{\mathrm{i}\sqrt{\pi}}{\mathrm{2}}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{i}\right)\right)=\frac{\mathrm{i}\sqrt{\mathrm{2}\pi}}{\mathrm{4}}\left(\mathrm{1}+\mathrm{i}\right) \\ $$

Question Number 193314    Answers: 1   Comments: 0

a ,b, c > 0 & a^2 +b^2 +c^2 =3 prove that (((1+(3/(ab+bc+ca)) )^((a+b+c)^2 ) ))^(1/3) ≤(1+(a/b))(1+(b/c))(1+(c/a))

$$ \\ $$$$\boldsymbol{{a}}\:,\boldsymbol{{b}},\:\boldsymbol{{c}}\:\:>\:\mathrm{0}\:\&\:\:\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} =\mathrm{3}\:\boldsymbol{{prove}}\:\boldsymbol{{that}}\: \\ $$$$\sqrt[{\mathrm{3}}]{\left(\mathrm{1}+\frac{\mathrm{3}}{\boldsymbol{{ab}}+\boldsymbol{{bc}}+\boldsymbol{{ca}}}\:\right)^{\left(\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} } \:}\leqslant\left(\mathrm{1}+\frac{\boldsymbol{{a}}}{\boldsymbol{{b}}}\right)\left(\mathrm{1}+\frac{\boldsymbol{{b}}}{\boldsymbol{{c}}}\right)\left(\mathrm{1}+\frac{\boldsymbol{{c}}}{\boldsymbol{{a}}}\right) \\ $$

Question Number 193309    Answers: 0   Comments: 1

Question Number 193307    Answers: 0   Comments: 0

Question Number 193296    Answers: 2   Comments: 1

If a^2 + b^2 + c^2 = 16, x^2 + y^2 + z^2 = 25 and ax + by + cz = 20 then what is the value of ((a + b + c)/(x + y + z)) ?

$$\mathrm{If}\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:=\:\mathrm{16},\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \:=\:\mathrm{25} \\ $$$$\mathrm{and}\:{ax}\:+\:{by}\:+\:{cz}\:=\:\mathrm{20}\:\mathrm{then}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\:\frac{{a}\:+\:{b}\:+\:{c}}{{x}\:+\:{y}\:+\:{z}}\:? \\ $$

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