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Question Number 221260 Answers: 1 Comments: 0

Question Number 221256 Answers: 0 Comments: 0

∫∫∫_( [0,1]) ((ln[(1+x)(1+y)(1+z])/(1 + xyz)) dxdydz

$$ \\ $$$$\:\:\int\int\int_{\:\left[\mathrm{0},\mathrm{1}\right]} \:\frac{\mathrm{ln}\left[\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{y}\right)\left(\mathrm{1}+{z}\right]\right.}{\mathrm{1}\:+\:{xyz}}\:{dxdydz}\:\:\:\: \\ $$$$ \\ $$

Question Number 221255 Answers: 1 Comments: 3

Let X be a point inside a square ABCD, such that XA = 10 cm, XB = 6 cm and XC = 14 cm. Find the area of the square.

$$\mathrm{Let}\:\mathrm{X}\:\mathrm{be}\:\mathrm{a}\:\mathrm{point}\:\mathrm{inside}\:\mathrm{a}\:\mathrm{square}\:\mathrm{ABCD},\: \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{XA}\:=\:\mathrm{10}\:\mathrm{cm},\:\mathrm{XB}\:=\:\mathrm{6}\:\mathrm{cm}\:\mathrm{and} \\ $$$$\mathrm{XC}\:=\:\mathrm{14}\:\mathrm{cm}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}. \\ $$

Question Number 221254 Answers: 0 Comments: 0

∫∫∫_([0,1]^3 ) (1/((1−xyz)(1−xy)(1−yz)(1−zx))) dxdydz

$$\:\:\int\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{3}} } \:\frac{\mathrm{1}}{\left(\mathrm{1}−{xyz}\right)\left(\mathrm{1}−{xy}\right)\left(\mathrm{1}−{yz}\right)\left(\mathrm{1}−{zx}\right)}\:{dxdydz}\:\:\:\: \\ $$$$ \\ $$

Question Number 221247 Answers: 1 Comments: 1

Question Number 221245 Answers: 1 Comments: 0

Which is greater ((√7)+1) or ((√3)+(√5))?????

$${Which}\:{is}\:{greater}\:\:\:\:\:\:\:\left(\sqrt{\mathrm{7}}+\mathrm{1}\right)\:\:{or}\:\:\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}}\right)????? \\ $$

Question Number 221239 Answers: 1 Comments: 0

2^x =4^y =8^z and ((1/(2x))+1(1/(4y))+(1/(6z)))=((24)/7) z=??

$$\mathrm{2}^{{x}} =\mathrm{4}^{{y}} =\mathrm{8}^{{z}} \:\:{and}\:\left(\frac{\mathrm{1}}{\mathrm{2}{x}}+\mathrm{1}\frac{\mathrm{1}}{\mathrm{4}{y}}+\frac{\mathrm{1}}{\mathrm{6}{z}}\right)=\frac{\mathrm{24}}{\mathrm{7}}\:\:\: \\ $$$${z}=?? \\ $$

Question Number 221238 Answers: 3 Comments: 0

a=5+2(√6) then {(√a)−(1/( (√a)))}=??

$${a}=\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\:{then}\:\left\{\sqrt{{a}}−\frac{\mathrm{1}}{\:\sqrt{{a}}}\right\}=?? \\ $$

Question Number 221234 Answers: 2 Comments: 1

let 0 ≤ a,b,c, ≤ 2 , and a + b + c = 3 Prove that; (3/2) ≤ (1/(a + 1)) + (1/(b + 1)) + (1/(c +1 )) ≤ ((11)/6)

$$ \\ $$$$\:\:\mathrm{let}\:\mathrm{0}\:\leqslant\:{a},{b},{c},\:\leqslant\:\mathrm{2}\:,\:\mathrm{and}\:{a}\:+\:{b}\:+\:{c}\:=\:\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Prove}\:\mathrm{that}; \\ $$$$\:\:\frac{\mathrm{3}}{\mathrm{2}}\:\leqslant\:\frac{\mathrm{1}}{{a}\:+\:\mathrm{1}}\:+\:\frac{\mathrm{1}}{{b}\:+\:\mathrm{1}}\:+\:\frac{\mathrm{1}}{{c}\:+\mathrm{1}\:}\:\leqslant\:\frac{\mathrm{11}}{\mathrm{6}} \\ $$$$ \\ $$

Question Number 221233 Answers: 1 Comments: 2

if a,b,c,d,e,f > 0 and abcdef = 1 , then (1/( (√(1 + ad)))) + (1/( (√(1 + be)))) + (1/( (√(1 + cf)))) ≤ (3/( (√2))) Profosed by Craciun Georghe

$$ \\ $$$$\:\mathrm{if}\:{a},{b},{c},{d},{e},{f}\:>\:\mathrm{0}\:\mathrm{and}\:{abcdef}\:=\:\mathrm{1}\:, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{then} \\ $$$$\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:+\:{ad}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:+\:{be}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:+\:{cf}}}\:\leqslant\:\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}\: \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Profosed}\:\mathrm{by}\:\mathrm{Craciun}\:\mathrm{Georghe} \\ $$

Question Number 221228 Answers: 1 Comments: 0

Σ_(n = 1) ^∞ Σ_(m = 1) ^∞ (1/((m^2 + n^2 )^2 ))

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\:\underset{{m}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\left({m}^{\mathrm{2}} \:+\:{n}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$ \\ $$

Question Number 221218 Answers: 0 Comments: 1

Question Number 221203 Answers: 1 Comments: 0

∫ tan(((1/x)/(sec(x)))) + ((1 − sec(x))/(sec(x))) dx

$$ \\ $$$$\:\:\:\:\:\int\:\mathrm{tan}\left(\frac{\frac{\mathrm{1}}{{x}}}{\mathrm{sec}\left({x}\right)}\right)\:+\:\frac{\mathrm{1}\:−\:\mathrm{sec}\left({x}\right)}{\mathrm{sec}\left({x}\right)}\:\mathrm{d}{x} \\ $$$$ \\ $$

Question Number 221196 Answers: 1 Comments: 0

Question Number 221195 Answers: 1 Comments: 1

Question Number 221188 Answers: 1 Comments: 2

Question Number 221185 Answers: 5 Comments: 0

Question Number 221184 Answers: 3 Comments: 2

Question Number 221179 Answers: 0 Comments: 0

Question Number 221170 Answers: 3 Comments: 1

Question Number 221180 Answers: 0 Comments: 0

Question Number 221168 Answers: 1 Comments: 0

Question Number 221154 Answers: 0 Comments: 1

f(x)= (x/(∣ x ∣ + 1)) f(f(f(f(x)))) =?

$$\:{f}\left({x}\right)=\:\frac{{x}}{\mid\:{x}\:\mid\:+\:\mathrm{1}} \\ $$$$\:\:{f}\left({f}\left({f}\left({f}\left({x}\right)\right)\right)\right)\:=? \\ $$

Question Number 221153 Answers: 1 Comments: 1

f(x)= (1/2^x ) + (1/3^x ) + (1/4^x ) + ... +(1/(4000^x )) f(2) + f(3) + f(4)+ ... =?

$$\:\:{f}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}^{{x}} }\:+\:\frac{\mathrm{1}}{\mathrm{3}^{{x}} }\:+\:\frac{\mathrm{1}}{\mathrm{4}^{{x}} }\:+\:...\:+\frac{\mathrm{1}}{\mathrm{4000}^{{x}} } \\ $$$$\:\:{f}\left(\mathrm{2}\right)\:+\:{f}\left(\mathrm{3}\right)\:+\:{f}\left(\mathrm{4}\right)+\:...\:=? \\ $$

Question Number 221151 Answers: 0 Comments: 1

((√(x^2 −x−(√(x^2 −x−(√(x^2 −x−(√(...))))))))/( ((x^2 (√(x ((x^2 (√(x...))))^(1/3) ))))^(1/3) )) = (3/4) ⇒ (2/x) =?

$$\:\frac{\sqrt{{x}^{\mathrm{2}} −{x}−\sqrt{{x}^{\mathrm{2}} −{x}−\sqrt{{x}^{\mathrm{2}} −{x}−\sqrt{...}}}}}{\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} \:\sqrt{{x}\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} \:\sqrt{{x}...}}}}}\:=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\:\Rightarrow\:\frac{\mathrm{2}}{{x}}\:=?\: \\ $$

Question Number 221135 Answers: 1 Comments: 2

South Korean Grade 12 math Prove log_a M^n =nlog_a M Using below: When M=a^x , log_a M=x When N=a^y , log_a N=y MN=a^x ×a^y =a^(x+y) So, log_a (MN)=log_a (a^(x+y) )=x+y=log_a M+log_a N

$$\mathrm{South}\:\mathrm{Korean}\:\mathrm{Grade}\:\mathrm{12}\:\mathrm{math} \\ $$$$\mathrm{Prove}\:\mathrm{log}_{{a}} {M}^{{n}} ={n}\mathrm{log}_{{a}} {M} \\ $$$$\mathrm{Using}\:\mathrm{below}: \\ $$$$\mathrm{When}\:{M}={a}^{{x}} ,\:\mathrm{log}_{{a}} {M}={x} \\ $$$$\mathrm{When}\:{N}={a}^{{y}} ,\:\mathrm{log}_{{a}} {N}={y} \\ $$$${MN}={a}^{{x}} ×{a}^{{y}} ={a}^{{x}+{y}} \\ $$$$\mathrm{So},\:\mathrm{log}_{{a}} \left({MN}\right)=\mathrm{log}_{{a}} \left({a}^{{x}+{y}} \right)={x}+{y}=\mathrm{log}_{{a}} {M}+\mathrm{log}_{{a}} {N} \\ $$

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