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Question Number 227732 Answers: 0 Comments: 0

Question Number 227731 Answers: 0 Comments: 0

Question Number 227730 Answers: 0 Comments: 0

Question Number 227715 Answers: 1 Comments: 0

x > 6 y > 2 (√(x^2 − 36)) + (√(y^2 − 4)) = 6 min { x + y } = ?

$$\mathrm{x}\:>\:\mathrm{6} \\ $$$$\mathrm{y}\:>\:\mathrm{2} \\ $$$$\sqrt{\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{36}}\:\:+\:\:\sqrt{\mathrm{y}^{\mathrm{2}} \:−\:\mathrm{4}}\:\:=\:\:\mathrm{6} \\ $$$$\mathrm{min}\:\left\{\:\mathrm{x}\:+\:\mathrm{y}\:\right\}\:=\:? \\ $$

Question Number 227709 Answers: 0 Comments: 0

Question Number 227710 Answers: 0 Comments: 2

Question Number 227694 Answers: 2 Comments: 0

max x^(1/x) =?

$${max}\:{x}^{\frac{\mathrm{1}}{{x}}} =? \\ $$

Question Number 227692 Answers: 3 Comments: 1

Question Number 227682 Answers: 0 Comments: 1

Help please I = ∫_0 ^( π) cos(x + cos(x)) dx

$$\mathrm{Help}\:\mathrm{please} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\:\pi} \mathrm{cos}\left({x}\:+\:\mathrm{cos}\left({x}\right)\right)\:{dx} \\ $$

Question Number 227687 Answers: 2 Comments: 4

Question Number 227677 Answers: 1 Comments: 0

z = 2−3i (1−z) ∙ ((1 + z^2 )/4) = ?

$$\mathrm{z}\:=\:\mathrm{2}−\mathrm{3i} \\ $$$$\left(\mathrm{1}−\mathrm{z}\right)\:\centerdot\:\frac{\mathrm{1}\:+\:\mathrm{z}^{\mathrm{2}} }{\mathrm{4}}\:=\:? \\ $$

Question Number 227673 Answers: 1 Comments: 0

Question Number 227670 Answers: 0 Comments: 5

Question Number 227667 Answers: 1 Comments: 0

sin7xcos5x−cos7xsin5x = (1/2) x = ?

$$\mathrm{sin7xcos5x}−\mathrm{cos7xsin5x}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{x}\:=\:? \\ $$

Question Number 227656 Answers: 1 Comments: 0

Question Number 227652 Answers: 2 Comments: 0

if (a/(y+z))=(b/(z+x))=(c/(x+y)) prove ((a(b−c))/(y^2 −z^2 ))=((b(c−a))/(z^2 −x^2 ))=((c(a−b))/(x^2 −y^2 ))

$${if} \\ $$$$\frac{{a}}{{y}+{z}}=\frac{{b}}{{z}+{x}}=\frac{{c}}{{x}+{y}}\: \\ $$$${prove} \\ $$$$\frac{{a}\left({b}−{c}\right)}{{y}^{\mathrm{2}} −{z}^{\mathrm{2}} }=\frac{{b}\left({c}−{a}\right)}{{z}^{\mathrm{2}} −{x}^{\mathrm{2}} }=\frac{{c}\left({a}−{b}\right)}{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } \\ $$

Question Number 227651 Answers: 1 Comments: 0

log_5 x = 25 x = ?

$$\mathrm{log}_{\mathrm{5}} \:\mathrm{x}\:=\:\mathrm{25} \\ $$$$\mathrm{x}\:=\:? \\ $$

Question Number 227649 Answers: 1 Comments: 0

25^(log_5 (√2) + 1) = ?

$$\mathrm{25}^{\boldsymbol{\mathrm{log}}_{\mathrm{5}} \:\sqrt{\mathrm{2}}\:+\:\mathrm{1}} \:\:\:=\:\:\:? \\ $$

Question Number 227635 Answers: 0 Comments: 0

prove: 8Σ_(n=0) ^∞ (1/((2n−1)sinh((2n−1)π)))=ln2

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{prove}:\:\mathrm{8}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right)\mathrm{sinh}\left(\left(\mathrm{2}{n}−\mathrm{1}\right)\pi\right)}=\mathrm{ln2} \\ $$

Question Number 227634 Answers: 0 Comments: 0

prove: p>1,n≥2 (1/n^p )<(1/(p−1))∙[(1/((n−1)^(p− 1) ))−(1/n^(p−1) )]

$$\mathrm{prove}:\:\:\:\:\:\:\:\:\:{p}>\mathrm{1},{n}\geqslant\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{{n}^{{p}} }<\frac{\mathrm{1}}{{p}−\mathrm{1}}\centerdot\left[\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{{p}−\:\mathrm{1}} }−\frac{\mathrm{1}}{{n}^{{p}−\mathrm{1}} }\right] \\ $$

Question Number 227633 Answers: 1 Comments: 0

(3/(log_2 23!)) + (3/(log_3 23!)) + (3/(log_4 23!)) +...+ (3/(log_(22) 23!)) + (3/(log_(23) 23!)) = ?

$$\frac{\mathrm{3}}{\mathrm{log}_{\mathrm{2}} \mathrm{23}!}\:+\:\frac{\mathrm{3}}{\mathrm{log}_{\mathrm{3}} \mathrm{23}!}\:+\:\frac{\mathrm{3}}{\mathrm{log}_{\mathrm{4}} \mathrm{23}!}\:+...+\:\frac{\mathrm{3}}{\mathrm{log}_{\mathrm{22}} \mathrm{23}!}\:+\:\frac{\mathrm{3}}{\mathrm{log}_{\mathrm{23}} \mathrm{23}!}\:=\:? \\ $$

Question Number 227632 Answers: 1 Comments: 0

If 4a^2 + 9b^2 = 13ab Find ((2lg(2a + 3b)−lg25)/(5lg(ab))) = ?

$$\mathrm{If}\:\:\:\mathrm{4a}^{\mathrm{2}} \:+\:\mathrm{9b}^{\mathrm{2}} \:=\:\mathrm{13ab} \\ $$$$\mathrm{Find}\:\:\:\frac{\mathrm{2}\boldsymbol{\mathrm{lg}}\left(\mathrm{2a}\:+\:\mathrm{3b}\right)−\boldsymbol{\mathrm{lg}}\mathrm{25}}{\mathrm{5}\boldsymbol{\mathrm{lg}}\left(\mathrm{ab}\right)}\:=\:? \\ $$

Question Number 227629 Answers: 1 Comments: 0

Let the equations on x (a,b,c∈R; a≠0) ax^2 +bx+c=0 w/ roots r and s and, x^2 +bx+ac=0 w/ roots m and n Prove that: r=(m/a) and s=(n/a)

$${Let}\:{the}\:{equations}\:{on}\:{x}\:\left({a},{b},{c}\in\mathbb{R};\:{a}\neq\mathrm{0}\right) \\ $$$${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:{w}/\:{roots}\:{r}\:{and}\:{s} \\ $$$$\:{and},\:{x}^{\mathrm{2}} +{bx}+{ac}=\mathrm{0}\:{w}/\:{roots}\:{m}\:{and}\:{n} \\ $$$${Prove}\:{that}:\:{r}=\frac{{m}}{{a}}\:{and}\:{s}=\frac{{n}}{{a}} \\ $$

Question Number 227612 Answers: 0 Comments: 0

m_− R (R/m_− ) ⇔

$$ \\ $$$$\:\:\:\:\underset{−} {{m}}\: {R} \\ $$$$\:\:\:\:\:\:\: \:\:\frac{{R}}{\underset{−} {{m}}}\:\:\:\:\Leftrightarrow \: \\ $$

Question Number 227605 Answers: 0 Comments: 3

the two roots of eq x^2 +2ax−2b^2 =0 are α&β a,b are rational but a^2 +b^2 is not complete square. make a quad eq whose one root is α+β+(√(α^2 +β^2 ))

$${the}\:{two}\:{roots}\:{of}\:{eq}\:{x}^{\mathrm{2}} +\mathrm{2}{ax}−\mathrm{2}{b}^{\mathrm{2}} =\mathrm{0}\:{are} \\ $$$$\alpha\&\beta \\ $$$${a},{b}\:{are}\:{rational}\:{but}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:{is}\:{not}\:{complete} \\ $$$${square}.\:{make}\:{a}\:{quad}\:{eq}\:{whose}\:{one}\:{root}\:{is}\: \\ $$$$\alpha+\beta+\sqrt{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} } \\ $$

Question Number 227607 Answers: 2 Comments: 0

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