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Question Number 109605 Answers: 0 Comments: 3

Question Number 109595 Answers: 1 Comments: 1

For any Real numbers x,y and z, if (x+y+z)=2, then prove that xyz≥8(1−x)(1−y)(1−z)

$${For}\:{any}\:{Real}\:{numbers}\:{x},{y}\:{and}\:{z}, \\ $$$$\:{if}\:\:\left({x}+{y}+{z}\right)=\mathrm{2},\:{then}\:{prove}\:{that} \\ $$$$\:\:\:\:\:\:\:\:{xyz}\geqslant\mathrm{8}\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)\left(\mathrm{1}−{z}\right) \\ $$

Question Number 109591 Answers: 1 Comments: 3

An isosceles AEF is inscribed into a square ABCD such that pointE is on side BC,point F is on side CDand AE=EF. Knownthat tanAEF^() =2.Find tanFEC^()

$$\mathrm{An}\:\mathrm{isosceles}\:\mathrm{AEF}\:\mathrm{is}\:\mathrm{inscribed}\:\mathrm{into}\:\mathrm{a} \\ $$$$\mathrm{square}\:\mathrm{ABCD}\:\mathrm{such}\:\mathrm{that}\:\mathrm{pointE}\:\mathrm{is}\:\mathrm{on} \\ $$$$\mathrm{side}\:\mathrm{BC},\mathrm{point}\:\mathrm{F}\:\mathrm{is}\:\mathrm{on}\:\mathrm{side}\:\mathrm{CDand}\:\mathrm{AE}=\mathrm{EF}. \\ $$$$\mathrm{Knownthat}\:\mathrm{tan}\widehat {\mathrm{AEF}}=\mathrm{2}.\mathrm{Find}\:\mathrm{tan}\widehat {\mathrm{FEC}} \\ $$

Question Number 109585 Answers: 1 Comments: 0

If (xy+yz+zx)=1, then prove that (x/(1−x^2 ))+(y/(1−y^2 ))+(z/(1−z^2 ))=((4xyz)/((1−x^2 )(1−y^2 )(1−z^2 )))

$${If}\:\left({xy}+{yz}+{zx}\right)=\mathrm{1},\:{then}\:{prove}\:{that} \\ $$$$\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{{y}}{\mathrm{1}−{y}^{\mathrm{2}} }+\frac{{z}}{\mathrm{1}−{z}^{\mathrm{2}} }=\frac{\mathrm{4}{xyz}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}−{y}^{\mathrm{2}} \right)\left(\mathrm{1}−{z}^{\mathrm{2}} \right)} \\ $$

Question Number 109584 Answers: 3 Comments: 1

((−♭o♭−)/(hans)) (1)(√(x−(√(x−(1/4))))) ≥ (1/4) (2)∣a^→ ∣ = 1, ∣b^→ ∣ = 2 , ∣c^→ ∣=3 , ∠(a^→ ,b^→ )=90° ∠(b^→ ,c^→ )=60° , ∠(a^→ ,c^→ )=120° , then ∣a^→ +b^→ −c^→ ∣=? (3) { ((x^(log _3 (y)) +y^(log _3 (x)) =18)),((log _3 (x)+log _3 (y)=3)) :} . Find the solution

$$\:\:\:\frac{−\flat{o}\flat−}{{hans}} \\ $$$$\left(\mathrm{1}\right)\sqrt{{x}−\sqrt{{x}−\frac{\mathrm{1}}{\mathrm{4}}}}\:\geqslant\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\left(\mathrm{2}\right)\mid\overset{\rightarrow} {{a}}\mid\:=\:\mathrm{1},\:\mid\overset{\rightarrow} {{b}}\mid\:=\:\mathrm{2}\:,\:\mid\overset{\rightarrow} {{c}}\mid=\mathrm{3}\:,\:\angle\left(\overset{\rightarrow} {{a}},\overset{\rightarrow} {{b}}\right)=\mathrm{90}° \\ $$$$\:\:\:\:\:\:\:\:\angle\left(\overset{\rightarrow} {{b}},\overset{\rightarrow} {{c}}\right)=\mathrm{60}°\:,\:\angle\left(\overset{\rightarrow} {{a}},\overset{\rightarrow} {{c}}\right)=\mathrm{120}°\:,\:{then}\: \\ $$$$\:\:\:\:\:\:\mid\overset{\rightarrow} {{a}}+\overset{\rightarrow} {{b}}−\overset{\rightarrow} {{c}}\mid=? \\ $$$$\left(\mathrm{3}\right)\begin{cases}{{x}^{\mathrm{log}\:_{\mathrm{3}} \left({y}\right)} +{y}^{\mathrm{log}\:_{\mathrm{3}} \left({x}\right)} =\mathrm{18}}\\{\mathrm{log}\:_{\mathrm{3}} \left({x}\right)+\mathrm{log}\:_{\mathrm{3}} \left({y}\right)=\mathrm{3}}\end{cases}\:.\:{Find}\:{the}\:{solution} \\ $$$$\:\:\: \\ $$

Question Number 109577 Answers: 4 Comments: 1

Given x^4 +x^2 y^2 +y^4 =133 and x^2 −xy+y^2 =7 then what is the value of xy ?

$$\:\:\:\mathrm{G}{iven}\:{x}^{\mathrm{4}} +{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{4}} =\mathrm{133} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{and}\:{x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} =\mathrm{7} \\ $$$$\:\:{then}\:{what}\:{is}\:{the}\:{value}\:{of}\:{xy}\:? \\ $$

Question Number 109574 Answers: 0 Comments: 6

Question Number 109569 Answers: 1 Comments: 0

Question Number 109553 Answers: 1 Comments: 2

Question Number 109546 Answers: 1 Comments: 2

If f(x) continue in [ 1,30] and ∫_6 ^(30) f(x)dx = 30, then ∫_1 ^9 f(3y+3)dy = __

$${If}\:{f}\left({x}\right)\:{continue}\:{in}\:\left[\:\mathrm{1},\mathrm{30}\right]\:{and}\: \\ $$$$\underset{\mathrm{6}} {\overset{\mathrm{30}} {\int}}{f}\left({x}\right){dx}\:=\:\mathrm{30},\:{then}\:\underset{\mathrm{1}} {\overset{\mathrm{9}} {\int}}{f}\left(\mathrm{3}{y}+\mathrm{3}\right){dy}\:=\:\_\_ \\ $$

Question Number 109544 Answers: 2 Comments: 1

Exclude m and n from the equalities: a=m+n,b^3 =m^3 +n^3 ,c^5 =m^5 +n^5

$$\mathrm{Exclude}\:\mathrm{m}\:\mathrm{and}\:\mathrm{n}\:\mathrm{from}\:\mathrm{the}\:\mathrm{equalities}: \\ $$$$\mathrm{a}=\mathrm{m}+\mathrm{n},\mathrm{b}^{\mathrm{3}} =\mathrm{m}^{\mathrm{3}} +\mathrm{n}^{\mathrm{3}} ,\mathrm{c}^{\mathrm{5}} =\mathrm{m}^{\mathrm{5}} +\mathrm{n}^{\mathrm{5}} \\ $$

Question Number 109540 Answers: 1 Comments: 1

Question Number 109516 Answers: 3 Comments: 0

cos (1−i)=a+ib Find a, b.

$$\mathrm{cos}\:\left(\mathrm{1}−{i}\right)={a}+{ib} \\ $$$${Find}\:\:{a},\:{b}. \\ $$

Question Number 109506 Answers: 0 Comments: 0

Question Number 109509 Answers: 4 Comments: 0

((bemath)/(Σ_(i=cooll) ^(nice) (joss)_i )) ∫ ((x^2 dx)/( (√(x^2 +25))))

$$\:\:\frac{{bemath}}{\underset{{i}={cooll}} {\overset{{nice}} {\sum}}\left({joss}\right)_{{i}} }\: \\ $$$$ \\ $$$$\int\:\frac{{x}^{\mathrm{2}} \:{dx}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}} \\ $$

Question Number 109508 Answers: 2 Comments: 1

Question Number 109500 Answers: 3 Comments: 0

Given { ((a^2 +ab+bc+ac=a+c)),((b^2 +ab+bc+ac=b+a)),((c^2 +ab+bc+ac=c+b)) :} find the value of a+b+c

$${Given}\:\begin{cases}{{a}^{\mathrm{2}} +{ab}+{bc}+{ac}={a}+{c}}\\{{b}^{\mathrm{2}} +{ab}+{bc}+{ac}={b}+{a}}\\{{c}^{\mathrm{2}} +{ab}+{bc}+{ac}={c}+{b}}\end{cases} \\ $$$${find}\:{the}\:{value}\:{of}\:{a}+{b}+{c}\: \\ $$

Question Number 109497 Answers: 1 Comments: 0

Question Number 109495 Answers: 4 Comments: 0

1) ∫_0 ^(Π/2) sin x∙sin 2x∙sin 3x∙dx = ? 2) ∫_0 ^(1/2) arcsin x∙dx= ?

$$\left.\mathrm{1}\right)\:\:\:\:\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \mathrm{sin}\:{x}\centerdot\mathrm{sin}\:\mathrm{2}{x}\centerdot\mathrm{sin}\:\mathrm{3}{x}\centerdot{dx}\:=\:? \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{arcsin}\:{x}\centerdot{dx}=\:? \\ $$

Question Number 109494 Answers: 2 Comments: 0

Question Number 109493 Answers: 1 Comments: 0

Question Number 109489 Answers: 2 Comments: 0

(1) sin (2x)−cos (2x)−sin (x)+cos (x)=0 (2)lim_(x→0) ((e^x −e^(−x) )/(sin x)) (3)lim_(x→−1) ∣x+1∣ sin (x+1)

$$\left(\mathrm{1}\right)\:\mathrm{sin}\:\left(\mathrm{2}{x}\right)−\mathrm{cos}\:\left(\mathrm{2}{x}\right)−\mathrm{sin}\:\left({x}\right)+\mathrm{cos}\:\left({x}\right)=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{sin}\:{x}} \\ $$$$\left(\mathrm{3}\right)\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\mid{x}+\mathrm{1}\mid\:\mathrm{sin}\:\left({x}+\mathrm{1}\right) \\ $$$$ \\ $$

Question Number 109488 Answers: 0 Comments: 0

How many are the permutations of 1 − a little rubik′s cube with 4 squares by side 2 − a classical one with 9 squares by side

$${How}\:{many}\:{are}\:{the}\:{permutations}\:{of} \\ $$$$\mathrm{1}\:−\:{a}\:{little}\:{rubik}'{s}\:{cube}\:{with}\:\mathrm{4}\:{squares}\:{by}\:{side} \\ $$$$\mathrm{2}\:−\:{a}\:{classical}\:{one}\:{with}\:\mathrm{9}\:{squares}\:{by}\:{side} \\ $$

Question Number 109485 Answers: 1 Comments: 0

If f(x)=ax^2 +bx+c, g(x)= −ax^2 +bx+c where ac ≠ 0, then f(x)g(x)=0 has

$$\mathrm{If}\:{f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c},\:{g}\left({x}\right)=\:−{ax}^{\mathrm{2}} +{bx}+{c} \\ $$$$\mathrm{where}\:{ac}\:\neq\:\mathrm{0},\:\mathrm{then}\:{f}\left({x}\right){g}\left({x}\right)=\mathrm{0}\:\mathrm{has} \\ $$

Question Number 109483 Answers: 1 Comments: 2

Question Number 109472 Answers: 4 Comments: 0

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