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Question Number 188730 Answers: 2 Comments: 2

let p(x) = (5/3)−6x−9x^2 and Q(y) = −4y^2 −4y+((13)/2) if there exist unique pair of real number (x,y) such that p(x)×Q(y) = 20 then find the value 6x+10y = ?

$$\:\:\mathrm{let}\:{p}\left({x}\right)\:=\:\frac{\mathrm{5}}{\mathrm{3}}−\mathrm{6}{x}−\mathrm{9}{x}^{\mathrm{2}} \:\mathrm{and}\:{Q}\left({y}\right)\:=\:−\mathrm{4}{y}^{\mathrm{2}} −\mathrm{4}{y}+\frac{\mathrm{13}}{\mathrm{2}} \\ $$$$\mathrm{if}\:\mathrm{there}\:\mathrm{exist}\:\mathrm{unique}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{real}\:\mathrm{number} \\ $$$$\:\left({x},{y}\right)\:\mathrm{such}\:\mathrm{that}\:{p}\left({x}\right)×{Q}\left({y}\right)\:=\:\mathrm{20}\:\mathrm{then} \\ $$$$\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{6}{x}+\mathrm{10}{y}\:=\:? \\ $$

Question Number 188657 Answers: 3 Comments: 0

Question Number 188351 Answers: 2 Comments: 0

(√x)+(√(3x−2))=x^2 +1

$$\sqrt{{x}}+\sqrt{\mathrm{3}{x}−\mathrm{2}}={x}^{\mathrm{2}} +\mathrm{1} \\ $$

Question Number 188218 Answers: 1 Comments: 0

Question Number 188203 Answers: 1 Comments: 0

Question Number 188107 Answers: 2 Comments: 0

Question Number 187835 Answers: 5 Comments: 1

Question Number 187759 Answers: 0 Comments: 0

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Question Number 187713 Answers: 1 Comments: 1

Question Number 187662 Answers: 1 Comments: 0

Question Number 187651 Answers: 2 Comments: 0

Find the range of this function x^2 −13x + 36 = 0 Help!

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\mathrm{this}\:\mathrm{function} \\ $$$$\mathrm{x}^{\mathrm{2}} \:−\mathrm{13x}\:+\:\mathrm{36}\:=\:\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$

Question Number 187639 Answers: 2 Comments: 1

Question Number 187568 Answers: 0 Comments: 1

determiner la vitesse au[ponr A le temps mis par la balle pour passer de Oa E v_0 =5m/s

$${determiner}\:{la}\:{vitesse}\:{au}\left[{ponr}\:{A}\right. \\ $$$${le}\:{temps}\:{mis}\:{par}\:{la}\:{balle}\:{pour} \\ $$$${passer}\:{de}\:{Oa}\:{E}\:{v}_{\mathrm{0}} =\mathrm{5}{m}/{s} \\ $$

Question Number 187504 Answers: 0 Comments: 2

Question Number 187498 Answers: 1 Comments: 0

Question Number 187433 Answers: 1 Comments: 0

Prove that ▽×▽φ=0. If φ is assumed to be continous and having continous first and second order derivatives. M.m

$$\mathrm{Prove}\:\mathrm{that}\:\bigtriangledown×\bigtriangledown\phi=\mathrm{0}.\:\mathrm{If}\:\phi\:\mathrm{is} \\ $$$$\mathrm{assumed}\:\mathrm{to}\:\mathrm{be}\:\mathrm{continous}\:\mathrm{and} \\ $$$$\mathrm{having}\:\mathrm{continous}\:\mathrm{first}\:\mathrm{and}\:\mathrm{second} \\ $$$$\mathrm{order}\:\mathrm{derivatives}. \\ $$$$ \\ $$$$\mathrm{M}.\mathrm{m} \\ $$

Question Number 187155 Answers: 2 Comments: 2

f(x)=x.e^(2x) =>f^((n)) (x)=

$${f}\left({x}\right)={x}.{e}^{\mathrm{2}{x}} \\ $$$$=>{f}^{\left({n}\right)} \left({x}\right)= \\ $$$$ \\ $$

Question Number 186962 Answers: 1 Comments: 0

Question Number 186662 Answers: 1 Comments: 0

Question Number 186636 Answers: 1 Comments: 2

x^2 +x+1=0 x^(92) =?

$${x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:{x}^{\mathrm{92}} =? \\ $$

Question Number 186446 Answers: 0 Comments: 0

Question Number 186437 Answers: 1 Comments: 2

∫_0 ^( 1) (1/( (√(x(√(x^2 (√(x^3 (√(x^4 +1)))))))) )) dx

$$ \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{1}}{\:\sqrt{{x}\sqrt{{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{3}} \sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}}}\:}\:{dx} \\ $$$$\: \\ $$

Question Number 186230 Answers: 2 Comments: 0

Question Number 186080 Answers: 0 Comments: 0

Question Number 186077 Answers: 1 Comments: 0

Prove that ▽•(∅A^− )=(▽∅)•A+∅(▽•A^− ). Help!

$$\mathrm{Prove}\:\mathrm{that}\: \\ $$$$\bigtriangledown\bullet\left(\varnothing\overset{−} {\mathrm{A}}\right)=\left(\bigtriangledown\varnothing\right)\bullet\mathrm{A}+\varnothing\left(\bigtriangledown\bullet\overset{−} {\mathrm{A}}\right). \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$

Question Number 186074 Answers: 1 Comments: 0

Prove that div(curlA^− )=0 Help!

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{div}\left(\mathrm{curl}\overset{−} {\mathrm{A}}\right)=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$

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