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Question Number 184609 Answers: 2 Comments: 0

Question Number 184607 Answers: 0 Comments: 3

solve { ((x^2 −xy+y^2 =16)),((y^2 −yz+z^2 =25)),((z^2 −zx+x^2 =49)) :}

$${solve} \\ $$$$\begin{cases}{{x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} =\mathrm{16}}\\{{y}^{\mathrm{2}} −{yz}+{z}^{\mathrm{2}} =\mathrm{25}}\\{{z}^{\mathrm{2}} −{zx}+{x}^{\mathrm{2}} =\mathrm{49}}\end{cases} \\ $$

Question Number 184602 Answers: 1 Comments: 1

Question Number 184600 Answers: 0 Comments: 3

Question Number 184594 Answers: 1 Comments: 0

If f(x)=ln∣a+(1/(1−x))∣+b is an odd function, then find the value of a, b.

$$\mathrm{If}\:{f}\left({x}\right)=\mathrm{ln}\mid{a}+\frac{\mathrm{1}}{\mathrm{1}−{x}}\mid+{b}\:\mathrm{is}\:\mathrm{an}\:\mathrm{odd}\:\mathrm{function}, \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{a},\:{b}. \\ $$

Question Number 184590 Answers: 0 Comments: 0

Question Number 184574 Answers: 0 Comments: 5

Test whether this is Convergent or Divergent Σ_(n=0) ^∞ (−1)^n ((n!x^n )/(5n)) Help!

$$\mathrm{Test}\:\mathrm{whether}\:\mathrm{this}\:\mathrm{is}\:\mathrm{Convergent}\:\mathrm{or} \\ $$$$\mathrm{Divergent} \\ $$$$\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}} \frac{\mathrm{n}!\mathrm{x}^{\mathrm{n}} }{\mathrm{5n}} \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$

Question Number 184573 Answers: 4 Comments: 2

Question Number 184567 Answers: 1 Comments: 0

Resoudre dans Z^+ (√a) +(√b) =z (a,b,z)∈N^3 a,b ?

$${Resoudre}\:{dans}\:\mathbb{Z}^{+} \\ $$$$\sqrt{{a}}\:\:+\sqrt{{b}}\:={z}\:\:\:\:\left({a},{b},{z}\right)\in\mathbb{N}^{\mathrm{3}} \\ $$$${a},{b}\:? \\ $$

Question Number 184557 Answers: 1 Comments: 1

Determiner 1•AB, BC AC en fonction de r 2• ∡CBA ; ∡BAC ;et ∡BCA

$${Determiner} \\ $$$$\mathrm{1}\bullet\mathrm{AB},\:\:\mathrm{BC}\:\:\mathrm{AC}\:\mathrm{en}\:\mathrm{fonction}\:\mathrm{de}\:\boldsymbol{\mathrm{r}} \\ $$$$\mathrm{2}\bullet\:\:\measuredangle\mathrm{CBA}\:;\:\:\measuredangle\mathrm{BAC}\:;{et}\:\measuredangle\mathrm{BCA} \\ $$

Question Number 184555 Answers: 3 Comments: 6

Suppose that the sum of the square of complex numbers x or y is 7 , and the sum of their cubes is 10. Find the largest true value of the sum x+y that satisfies these conditions. A)4 B)5 C)6 D)7 E)8

$$\mathrm{Suppose}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\:\mathrm{of} \\ $$$$\mathrm{complex}\:\mathrm{numbers}\:\:\boldsymbol{\mathrm{x}}\:\mathrm{or}\:\boldsymbol{\mathrm{y}}\:\mathrm{is}\:\mathrm{7}\:,\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{their}\:\mathrm{cubes}\:\mathrm{is}\:\mathrm{10}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{largest} \\ $$$$\mathrm{true}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sum}\:\:\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}\:\:\mathrm{that}\:\mathrm{satisfies} \\ $$$$\mathrm{these}\:\mathrm{conditions}. \\ $$$$\left.\mathrm{A}\left.\right)\left.\mathrm{4}\left.\:\left.\:\:\mathrm{B}\right)\mathrm{5}\:\:\:\mathrm{C}\right)\mathrm{6}\:\:\:\mathrm{D}\right)\mathrm{7}\:\:\:\mathrm{E}\right)\mathrm{8} \\ $$

Question Number 184552 Answers: 2 Comments: 0

I_1 =∫_0 ^∞ (((√(x+(√(x^2 +1))))/( (√(x^2 +1))))−((√2)/( (√x))))dx=? I_2 =∫_0 ^∞ (((√(x^2 +1))/( (√(x+(√(x^2 +1))))))−((√x)/( (√2))))dx=?

$${I}_{\mathrm{1}} =\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\frac{\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}−\frac{\sqrt{\mathrm{2}}}{\:\sqrt{{x}}}\right){dx}=? \\ $$$${I}_{\mathrm{2}} =\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{\:\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}}−\frac{\sqrt{{x}}}{\:\sqrt{\mathrm{2}}}\right){dx}=? \\ $$

Question Number 184551 Answers: 1 Comments: 0

Resoudre dans Z^+ x+y+(√(xy)) =39

$${Resoudre}\:{dans}\:\mathbb{Z}^{+} \\ $$$${x}+{y}+\sqrt{{xy}}\:\:\:\:=\mathrm{39} \\ $$

Question Number 184538 Answers: 1 Comments: 1

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Question Number 184535 Answers: 6 Comments: 0

If a, b>0 such that 2a+b=2, then find the minimum value of: 1) (4a^2 +1)(b^2 +1) 2) ((2a^2 −b+4)/(a+1))+((b^2 −2a−2)/(b+4))

$$\mathrm{If}\:{a},\:{b}>\mathrm{0}\:\mathrm{such}\:\mathrm{that}\:\mathrm{2}{a}+{b}=\mathrm{2}, \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}: \\ $$$$\left.\mathrm{1}\right)\:\:\left(\mathrm{4}{a}^{\mathrm{2}} +\mathrm{1}\right)\left({b}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right)\:\:\frac{\mathrm{2}{a}^{\mathrm{2}} −{b}+\mathrm{4}}{{a}+\mathrm{1}}+\frac{{b}^{\mathrm{2}} −\mathrm{2}{a}−\mathrm{2}}{{b}+\mathrm{4}} \\ $$

Question Number 184523 Answers: 2 Comments: 0

Question Number 184514 Answers: 1 Comments: 0

Question Number 184511 Answers: 0 Comments: 0

In a regular heptagon ABCDEFG : (√(2(AC)^2 −(AD)^2 )) − AD = 2 find BC.

$${In}\:{a}\:{regular}\:{heptagon}\:{ABCDEFG}\:: \\ $$$$\:\sqrt{\mathrm{2}\left({AC}\right)^{\mathrm{2}} −\left({AD}\right)^{\mathrm{2}} }\:−\:{AD}\:=\:\mathrm{2} \\ $$$$\:{find}\:{BC}.\: \\ $$

Question Number 184505 Answers: 1 Comments: 0

(((a+b+c)^5 −a^5 −b^5 −c^5 )/((a+b+c)^3 −a^3 −b^3 −c^3 )) = ? is there an easier way.

Question Number 184503 Answers: 1 Comments: 0

Question Number 184497 Answers: 1 Comments: 0

Question Number 184492 Answers: 1 Comments: 2

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Question Number 184478 Answers: 1 Comments: 0

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