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Question Number 186503 Answers: 0 Comments: 1

function of , f (x) = ax + ∣ x ∣ is one to one .find ” a ” .

$$ \\ $$$$\:\:\:{function}\:{of}\:,\:{f}\:\left({x}\right)\:=\:{ax}\:\:+\:\mid\:{x}\:\mid\:{is}\:\:{one}\:{to}\:{one} \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:.{find}\:\:\:\:''\:\:\:\:{a}\:\:\:\:''\:\:. \\ $$$$\:\: \\ $$

Question Number 186500 Answers: 1 Comments: 0

lim_(x→+∞) (√(x+(√(x+(√(x+(√x)))))))−(√x)=? please solution

$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}}−\sqrt{\mathrm{x}}=? \\ $$$$\mathrm{please}\:\mathrm{solution} \\ $$

Question Number 186492 Answers: 1 Comments: 1

Question Number 186489 Answers: 1 Comments: 0

Question Number 186486 Answers: 0 Comments: 6

Question propose par Migma −−−−−−−−−−− calcul de X △ABC ∡ACB=40^° AB^2 =AC^2 +BC^2 −2AC×BCcos 40 AC=15 ; AB=X+4 ; BC=10 (X+4)^2 =15^2 +10^2 −300×cos 40 =325−229,81334 (X+4)^2 =95,18666706 posons Z=X+4 Z^2 =9,756^2 ⇒X+4=9,756 X=5,756

$${Question}\:{propose}\:{par} \\ $$$${Migma} \\ $$$$−−−−−−−−−−− \\ $$$${calcul}\:{de}\:{X} \\ $$$$ \\ $$$$\bigtriangleup{ABC}\:\:\:\:\:\:\measuredangle{ACB}=\mathrm{40}^{°} \\ $$$$\mathrm{AB}^{\mathrm{2}} =\mathrm{AC}^{\mathrm{2}} +\mathrm{BC}^{\mathrm{2}} −\mathrm{2AC}×\mathrm{BCcos}\:\mathrm{40}\:\:\:\: \\ $$$$\mathrm{AC}=\mathrm{15}\:\:;\:\:\:\:\mathrm{AB}=\mathrm{X}+\mathrm{4}\:;\:\:\:\mathrm{BC}=\mathrm{10} \\ $$$$\left(\mathrm{X}+\mathrm{4}\right)^{\mathrm{2}} =\mathrm{15}^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} −\mathrm{300}×\mathrm{cos}\:\mathrm{40} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{325}−\mathrm{229},\mathrm{81334} \\ $$$$\left({X}+\mathrm{4}\right)^{\mathrm{2}} =\mathrm{95},\mathrm{18666706} \\ $$$${posons}\:\:\:{Z}={X}+\mathrm{4} \\ $$$$\:\:\:{Z}^{\mathrm{2}} =\mathrm{9},\mathrm{756}^{\mathrm{2}} \:\:\:\:\Rightarrow\mathrm{X}+\mathrm{4}=\mathrm{9},\mathrm{756} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{X}=\mathrm{5},\mathrm{756} \\ $$$$ \\ $$

Question Number 186482 Answers: 2 Comments: 0

Question Number 186481 Answers: 0 Comments: 0

Question Number 186476 Answers: 0 Comments: 0

Question Number 186473 Answers: 0 Comments: 0

A metallic cube is subjected to heating such that as the metal expands, the total surface area increases at rate of 6.25 cm^2 s^(−1) . Calculate the rate at which each side of the cube is increasing when the volume is 51.2 cm^3 .

$$\mathrm{A}\:\mathrm{metallic}\:\mathrm{cube}\:\mathrm{is}\:\mathrm{subjected}\:\mathrm{to} \\ $$$$\mathrm{heating}\:\mathrm{such}\:\mathrm{that}\:\mathrm{as}\:\mathrm{the}\:\mathrm{metal} \\ $$$$\mathrm{expands},\:\mathrm{the}\:\mathrm{total}\:\mathrm{surface}\:\mathrm{area} \\ $$$$\mathrm{increases}\:\mathrm{at}\:\mathrm{rate}\:\mathrm{of}\:\mathrm{6}.\mathrm{25}\:\mathrm{cm}^{\mathrm{2}} \mathrm{s}^{−\mathrm{1}} . \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{rate}\:\mathrm{at}\:\mathrm{which}\:\mathrm{each} \\ $$$$\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cube}\:\mathrm{is}\:\mathrm{increasing}\:\mathrm{when} \\ $$$$\mathrm{the}\:\mathrm{volume}\:\mathrm{is}\:\mathrm{51}.\mathrm{2}\:\mathrm{cm}^{\mathrm{3}} . \\ $$

Question Number 186468 Answers: 1 Comments: 0

Simplify ((1^2 ∙2!+2^2 ∙3!+3^2 ∙4!+∙∙∙+n^2 (n+1)!−2)/((n+1)!)) to n^2 +n−2

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Simplify} \\ $$$$\frac{\mathrm{1}^{\mathrm{2}} \centerdot\mathrm{2}!+\mathrm{2}^{\mathrm{2}} \centerdot\mathrm{3}!+\mathrm{3}^{\mathrm{2}} \centerdot\mathrm{4}!+\centerdot\centerdot\centerdot+{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)!−\mathrm{2}}{\left({n}+\mathrm{1}\right)!} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{to} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{n}^{\mathrm{2}} +\mathrm{n}−\mathrm{2} \\ $$

Question Number 186464 Answers: 1 Comments: 15