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Question Number 128112 Answers: 3 Comments: 0

1 + 2 + 3 + 4 + ..... + 100 = ?

$$\:\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}\:+\:\mathrm{4}\:+\:.....\:+\:\mathrm{100}\:=\:? \\ $$

Question Number 128110 Answers: 1 Comments: 0

∫_0 ^∞ ∫_0 ^∞ ((sinx sin(x+y))/(x(x+y)))dxdy

$$\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \frac{{sinx}\:{sin}\left({x}+{y}\right)}{{x}\left({x}+{y}\right)}{dxdy} \\ $$

Question Number 128093 Answers: 1 Comments: 0

Question Number 128083 Answers: 2 Comments: 0

((8 − i)/(3 − 2i)) If the expression above is rewritten in the form a + bi, where a and b are real numbers, what is the value of a? A. 2 B. (8/3) C. 3 D. ((11)/3)

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{8}\:−\:{i}}{\mathrm{3}\:−\:\mathrm{2}{i}} \\ $$$$\:\mathrm{If}\:\mathrm{the}\:\mathrm{expression}\:\mathrm{above}\:\mathrm{is}\:\mathrm{rewritten}\: \\ $$$$\:\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:{a}\:+\:{bi},\:\mathrm{where}\:{a}\:\mathrm{and}\:{b}\:\mathrm{are} \\ $$$$\:\mathrm{real}\:\mathrm{numbers},\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{a}? \\ $$$$\:\mathrm{A}.\:\mathrm{2} \\ $$$$\:\mathrm{B}.\:\frac{\mathrm{8}}{\mathrm{3}} \\ $$$$\:\mathrm{C}.\:\mathrm{3} \\ $$$$\:\mathrm{D}.\:\frac{\mathrm{11}}{\mathrm{3}} \\ $$

Question Number 128030 Answers: 1 Comments: 0

99 × 99 = 9801 999 × 999 = 998001 9999 × 9999 = 99980001 99999 × 99999 = ? 999999 × 999999 = ?

$$\:\mathrm{99}\:×\:\mathrm{99}\:=\:\mathrm{9801} \\ $$$$\:\mathrm{999}\:×\:\mathrm{999}\:=\:\mathrm{998001} \\ $$$$\:\mathrm{9999}\:×\:\mathrm{9999}\:=\:\mathrm{99980001} \\ $$$$\:\mathrm{99999}\:×\:\mathrm{99999}\:=\:? \\ $$$$\:\mathrm{999999}\:×\:\mathrm{999999}\:=\:? \\ $$

Question Number 128008 Answers: 1 Comments: 0

If 347.9823 = (3/P) + 4Q + 7R + (9/(10)) + (8/(100)) + (2/S) + (3/T) Then find the value of P + Q + R + S + T

$$\mathrm{If}\:\mathrm{347}.\mathrm{9823}\:=\:\frac{\mathrm{3}}{{P}}\:+\:\mathrm{4}{Q}\:+\:\mathrm{7}{R}\:+\:\frac{\mathrm{9}}{\mathrm{10}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\frac{\mathrm{8}}{\mathrm{100}}\:+\:\frac{\mathrm{2}}{{S}}\:+\:\frac{\mathrm{3}}{{T}} \\ $$$${Then}\:{find}\:{the}\:{value}\:{of}\: \\ $$$${P}\:+\:{Q}\:+\:{R}\:+\:{S}\:+\:{T} \\ $$

Question Number 128001 Answers: 1 Comments: 0

(x − a) (x − b) (x − c) ..... (x − z) = ?

$$\left(\mathrm{x}\:−\:\mathrm{a}\right)\:\left(\mathrm{x}\:−\:\mathrm{b}\right)\:\left(\mathrm{x}\:−\:\mathrm{c}\right)\:.....\:\left(\mathrm{x}\:−\:\mathrm{z}\right)\:=\:? \\ $$

Question Number 127974 Answers: 1 Comments: 1

θ^(••) +(g/l)sinθ=0 Exact form (May include elliptic integral)

$$\overset{\bullet\bullet} {\theta}+\frac{{g}}{{l}}{sin}\theta=\mathrm{0} \\ $$$${Exact}\:{form}\:\left({May}\:{include}\:{elliptic}\:{integral}\right) \\ $$

Question Number 127970 Answers: 1 Comments: 0

You have given a positive integer N. Calculate ∫_( 0) ^( ∞) ((e^(2πx^2 ) − 1)/(e^(2πx^2 ) + 1))((1/x) − (x/(N^2 − x^2 ))) dx

$$\mathrm{You}\:\mathrm{have}\:\mathrm{given}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{integer}\:{N}.\: \\ $$$$\mathrm{Calculate}\: \\ $$$$\int_{\:\mathrm{0}} ^{\:\infty} \frac{{e}^{\mathrm{2}\pi{x}^{\mathrm{2}} } \:−\:\mathrm{1}}{{e}^{\mathrm{2}\pi{x}^{\mathrm{2}} } \:+\:\mathrm{1}}\left(\frac{\mathrm{1}}{{x}}\:−\:\frac{{x}}{{N}^{\mathrm{2}} \:−\:{x}^{\mathrm{2}} }\right)\:{dx} \\ $$

Question Number 127933 Answers: 1 Comments: 0

(1/((1/(x − 3)) + (1/(x + 4)))) If x > 0 and x ≠ 3, which of the following is equivalent to the expresion above? A. 2x + 1 B. x^2 + x − 12 C. ((x^2 + x − 12)/(2x + 1)) D. ((2x + 1)/(x^2 + x −12))

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{{x}\:−\:\mathrm{3}}\:+\:\frac{\mathrm{1}}{{x}\:+\:\mathrm{4}}} \\ $$$$ \\ $$$$\mathrm{If}\:{x}\:>\:\mathrm{0}\:\mathrm{and}\:{x}\:\neq\:\mathrm{3},\:\mathrm{which}\:\mathrm{of}\:\mathrm{the}\:\:\: \\ $$$$\mathrm{following}\:\mathrm{is}\:\mathrm{equivalent}\:\mathrm{to}\:\mathrm{the}\: \\ $$$$\mathrm{expresion}\:\mathrm{above}? \\ $$$$\mathrm{A}.\:\:\mathrm{2}{x}\:+\:\mathrm{1} \\ $$$$\mathrm{B}.\:\:{x}^{\mathrm{2}} \:+\:{x}\:−\:\mathrm{12}\: \\ $$$$\mathrm{C}.\:\:\frac{{x}^{\mathrm{2}} \:+\:{x}\:−\:\mathrm{12}}{\mathrm{2}{x}\:+\:\mathrm{1}} \\ $$$$\mathrm{D}.\:\:\frac{\mathrm{2}{x}\:+\:\mathrm{1}}{{x}^{\mathrm{2}} \:+\:{x}\:−\mathrm{12}} \\ $$

Question Number 127876 Answers: 0 Comments: 0

Consider the sequence defined by: 0<u_0 <1 and ∀n∈N, u_(n+1) =u_n −u_n ^2 . 1. Show that the sequence (u_n ) converges. What is its limit ? 2. Show that the series with general term u_n ^2 converges. 3. Show that the series with general terms ln((u_(n+1) /u_n )) and u_n diverge.

Question Number 127859 Answers: 1 Comments: 0

Question Number 127793 Answers: 2 Comments: 1

Σ_(n=1) ^∞ (n/(n!))cos(((πn)/5))

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}}{{n}!}{cos}\left(\frac{\pi{n}}{\mathrm{5}}\right) \\ $$

Question Number 127771 Answers: 0 Comments: 3

(1/(1−(π^2 /(1+π^2 −((2π^2 )/(2+π^2 −((3π^2 )/(3+π^2 −((4π^2 )/(4+π^2 −((5π^2 )/(5+π^2 ....))))))))))))

$$\frac{\mathrm{1}}{\mathrm{1}−\frac{\pi^{\mathrm{2}} }{\mathrm{1}+\pi^{\mathrm{2}} −\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{2}+\pi^{\mathrm{2}} −\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{3}+\pi^{\mathrm{2}} −\frac{\mathrm{4}\pi^{\mathrm{2}} }{\mathrm{4}+\pi^{\mathrm{2}} −\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{5}+\pi^{\mathrm{2}} ....}}}}}} \\ $$

Question Number 127982 Answers: 0 Comments: 1

Some Values .. Σ_(n=−∞) ^∞ e^(−πn^2 ) =(π^(1/4) /(Γ((3/4)))) Σ_(n=−∞) ^∞ e^(−2πn^2 ) =(π^(1/4) /(Γ((3/4)))) (((6+4(√2)))^(1/4) /2) Σ_(n=−∞) ^∞ e^(−6πn^2 ) =(π^(1/4) /(Γ((3/4)))).((√((1)^(1/4) +(3)^(1/4) +(4)^(1/4) +(9)^(1/4) ))/( (√(1728)))) Any Idea to prove ?

$${Some}\:{Values}\:.. \\ $$$$\underset{{n}=−\infty} {\overset{\infty} {\sum}}{e}^{−\pi{n}^{\mathrm{2}} } =\frac{\pi^{\frac{\mathrm{1}}{\mathrm{4}}} }{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)} \\ $$$$\underset{{n}=−\infty} {\overset{\infty} {\sum}}{e}^{−\mathrm{2}\pi{n}^{\mathrm{2}} } =\frac{\pi^{\frac{\mathrm{1}}{\mathrm{4}}} }{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}\:\frac{\sqrt[{\mathrm{4}}]{\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$$\underset{{n}=−\infty} {\overset{\infty} {\sum}}{e}^{−\mathrm{6}\pi{n}^{\mathrm{2}} } =\frac{\pi^{\frac{\mathrm{1}}{\mathrm{4}}} }{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}.\frac{\sqrt{\sqrt[{\mathrm{4}}]{\mathrm{1}}+\sqrt[{\mathrm{4}}]{\mathrm{3}}+\sqrt[{\mathrm{4}}]{\mathrm{4}}+\sqrt[{\mathrm{4}}]{\mathrm{9}}}}{\:\sqrt{\mathrm{1728}}} \\ $$$${Any}\:{Idea}\:{to}\:{prove}\:? \\ $$

Question Number 127682 Answers: 1 Comments: 2

(π^2 /(1+(π^2 /(3−π^2 +((9π^2 )/(5−3π^2 +((25π^2 )/(7−5π^2 +((49π^( 2) )/(9−7π^2 +((81π^2 )/(11−9π^2 +((121π^2 )/(.....))))))))))))))

$$\frac{\pi^{\mathrm{2}} }{\mathrm{1}+\frac{\pi^{\mathrm{2}} }{\mathrm{3}−\pi^{\mathrm{2}} +\frac{\mathrm{9}\pi^{\mathrm{2}} }{\mathrm{5}−\mathrm{3}\pi^{\mathrm{2}} +\frac{\mathrm{25}\pi^{\mathrm{2}} }{\mathrm{7}−\mathrm{5}\pi^{\mathrm{2}} +\frac{\mathrm{49}\pi^{\:\mathrm{2}} }{\mathrm{9}−\mathrm{7}\pi^{\mathrm{2}} +\frac{\mathrm{81}\pi^{\mathrm{2}} }{\mathrm{11}−\mathrm{9}\pi^{\mathrm{2}} +\frac{\mathrm{121}\pi^{\mathrm{2}} }{.....}}}}}}} \\ $$

Question Number 127666 Answers: 0 Comments: 11

Have a great year all of you It is 12.20 am in India (GMT+5.30) (12.20)^T =20.21 💐🌅

$${Have}\:{a}\:{great}\:{year}\:{all}\:{of}\:{you} \\ $$$${It}\:{is}\:\mathrm{12}.\mathrm{20}\:{am}\:{in}\:{India}\:\left({GMT}+\mathrm{5}.\mathrm{30}\right) \\ $$$$\left(\mathrm{12}.\mathrm{20}\right)^{{T}} =\mathrm{20}.\mathrm{21} \\ $$$$ \\ $$💐🌅

Question Number 127644 Answers: 0 Comments: 2

(1/(1−(1/(2−((1/2)/((3/2)−((1/3)/((4/3)−((1/4)/((5/4)−((1/5)/((6/5)−...))))))))))))

$$\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}−\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\mathrm{3}}{\mathrm{2}}−\frac{\frac{\mathrm{1}}{\mathrm{3}}}{\frac{\mathrm{4}}{\mathrm{3}}−\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\frac{\mathrm{5}}{\mathrm{4}}−\frac{\frac{\mathrm{1}}{\mathrm{5}}}{\frac{\mathrm{6}}{\mathrm{5}}−...}}}}}} \\ $$

Question Number 127552 Answers: 0 Comments: 0