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Question Number 43805    Answers: 1   Comments: 1

In a properly shuffled deck, what is the probability that the last 10 cards are in order: A♥, 2♥, 3♥, 4♥, 5♥, 9♠, 10♠, J♠, K♠, Q♠

$$\mathrm{In}\:\mathrm{a}\:\mathrm{properly}\:\mathrm{shuffled}\:\mathrm{deck},\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{last}\:\mathrm{10}\:\mathrm{cards}\:\mathrm{are}\:\mathrm{in}\:\mathrm{order}: \\ $$$${A}\heartsuit,\:\mathrm{2}\heartsuit,\:\mathrm{3}\heartsuit,\:\mathrm{4}\heartsuit,\:\mathrm{5}\heartsuit,\:\mathrm{9}\spadesuit,\:\mathrm{10}\spadesuit,\:\mathrm{J}\spadesuit,\:\mathrm{K}\spadesuit,\:\mathrm{Q}\spadesuit \\ $$

Question Number 43804    Answers: 1   Comments: 0

solve for ε s(1−α)=(1−ε)σT^4

$${solve}\:{for}\:\epsilon \\ $$$$ \\ $$$${s}\left(\mathrm{1}−\alpha\right)=\left(\mathrm{1}−\epsilon\right)\sigma{T}^{\mathrm{4}} \\ $$

Question Number 43803    Answers: 1   Comments: 0

∞ ∫ (1/(x ln x))dx = 0

$$\infty \\ $$$$\int\:\:\:\:\frac{\mathrm{1}}{{x}\:\mathrm{l}{n}\:{x}}{dx}\:\:\:=\: \\ $$$$\mathrm{0} \\ $$

Question Number 43793    Answers: 2   Comments: 0

The value of ^3 (√(20+14(√2)))+^3 (√(20−14(√2))) is…

$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\:^{\mathrm{3}} \sqrt{\mathrm{20}+\mathrm{14}\sqrt{\mathrm{2}}}+^{\mathrm{3}} \sqrt{\mathrm{20}−\mathrm{14}\sqrt{\mathrm{2}}}\:\mathrm{is}\ldots \\ $$

Question Number 43792    Answers: 0   Comments: 0

For π≤θ<2π given P=(1/2)cos θ−(1/4)sin 2θ−(1/8)cos 3θ+(1/(16))sin 4θ+(1/(32))cos 5θ −(1/(64))sin 6θ−(1/(128))cos 7θ+… Q=1−(1/2)sin θ−(1/4)cos 2θ+(1/8)sin 3θ+(1/(16))cos 4θ−(1/(32))sin 5θ −(1/(64))cos 6θ+(1/(128))sin 7θ+... so (P/Q)=((2(√7))/7). After sin θ=−(m/n), where m and n prime relatif. The value m+n is…

$$\mathrm{For}\:\pi\leqslant\theta<\mathrm{2}\pi \\ $$$$\mathrm{given}\: \\ $$$$\mathrm{P}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\theta−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\mathrm{2}\theta−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cos}\:\mathrm{3}\theta+\frac{\mathrm{1}}{\mathrm{16}}\mathrm{sin}\:\mathrm{4}\theta+\frac{\mathrm{1}}{\mathrm{32}}\mathrm{cos}\:\mathrm{5}\theta \\ $$$$−\frac{\mathrm{1}}{\mathrm{64}}\mathrm{sin}\:\mathrm{6}\theta−\frac{\mathrm{1}}{\mathrm{128}}\mathrm{cos}\:\mathrm{7}\theta+\ldots \\ $$$$\mathrm{Q}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\theta−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\:\:\mathrm{2}\theta+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{sin}\:\:\mathrm{3}\theta+\frac{\mathrm{1}}{\mathrm{16}}\mathrm{cos}\:\:\mathrm{4}\theta−\frac{\mathrm{1}}{\mathrm{32}}\mathrm{sin}\:\:\mathrm{5}\theta \\ $$$$−\frac{\mathrm{1}}{\mathrm{64}}\mathrm{cos}\:\mathrm{6}\theta+\frac{\mathrm{1}}{\mathrm{128}}\mathrm{sin}\:\mathrm{7}\theta+... \\ $$$$\mathrm{so}\:\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{7}}.\:\mathrm{After}\:\mathrm{sin}\:\theta=−\frac{\mathrm{m}}{\mathrm{n}},\:\mathrm{where}\:\mathrm{m}\:\mathrm{and}\:\mathrm{n} \\ $$$$\mathrm{prime}\:\mathrm{relatif}.\:\mathrm{The}\:\mathrm{value}\:\mathrm{m}+\mathrm{n}\:\mathrm{is}\ldots \\ $$$$ \\ $$

Question Number 43789    Answers: 1   Comments: 0

plz solve it for me i am learning (((81)/(125)))^(1/3)

$$\:{plz}\:{solve}\:{it}\:{for}\:{me}\:{i}\:{am}\:{learning} \\ $$$$ \\ $$$$\:\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{81}}{\mathrm{125}}} \\ $$

Question Number 43787    Answers: 1   Comments: 0

The total number of numbers of not more than 20 digits that are formed by using the digits 0, 1, 2, 3 and 4 is

$$\mathrm{The}\:\mathrm{total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{numbers}\:\mathrm{of}\:\mathrm{not} \\ $$$$\mathrm{more}\:\mathrm{than}\:\mathrm{20}\:\mathrm{digits}\:\mathrm{that}\:\mathrm{are}\:\mathrm{formed} \\ $$$$\mathrm{by}\:\mathrm{using}\:\mathrm{the}\:\mathrm{digits}\:\:\mathrm{0},\:\mathrm{1},\:\mathrm{2},\:\mathrm{3}\:\mathrm{and}\:\mathrm{4}\:\mathrm{is} \\ $$

Question Number 43783    Answers: 1   Comments: 5

Question Number 43767    Answers: 0   Comments: 0

The chance of an event happening is the square of the chance of a second event but the odds against the first are the cube of the odds against the second. The chances of the events are

$$\mathrm{The}\:\mathrm{chance}\:\mathrm{of}\:\mathrm{an}\:\mathrm{event}\:\mathrm{happening}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{square}\:\mathrm{of}\:\mathrm{the}\:\mathrm{chance}\:\mathrm{of}\:\mathrm{a}\:\mathrm{second}\:\mathrm{event} \\ $$$$\mathrm{but}\:\mathrm{the}\:\mathrm{odds}\:\mathrm{against}\:\mathrm{the}\:\mathrm{first}\:\mathrm{are}\:\mathrm{the} \\ $$$$\mathrm{cube}\:\mathrm{of}\:\mathrm{the}\:\mathrm{odds}\:\mathrm{against}\:\mathrm{the}\:\mathrm{second}.\:\mathrm{The} \\ $$$$\mathrm{chances}\:\mathrm{of}\:\mathrm{the}\:\mathrm{events}\:\mathrm{are} \\ $$

Question Number 43766    Answers: 1   Comments: 0

If ∣a∣<1 and ∣b∣<1, then the sum of the series a(a+b)+a^2 (a^2 +b^2 )+a^3 (a^3 +b^3 )+... upto ∞ is

$$\mathrm{If}\:\mid{a}\mid<\mathrm{1}\:\mathrm{and}\:\mid{b}\mid<\mathrm{1},\:\mathrm{then}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{series}\:{a}\left({a}+{b}\right)+{a}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)+{a}^{\mathrm{3}} \left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right)+... \\ $$$$\mathrm{upto}\:\infty\:\mathrm{is} \\ $$

Question Number 43765    Answers: 1   Comments: 0

The sum of integers from 1 to 100 that are divisible by 2 or 5 is _____.

$$\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{integers}\:\mathrm{from}\:\mathrm{1}\:\mathrm{to}\:\mathrm{100}\:\mathrm{that} \\ $$$$\mathrm{are}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{2}\:\mathrm{or}\:\mathrm{5}\:\mathrm{is}\:\_\_\_\_\_. \\ $$

Question Number 43796    Answers: 1   Comments: 0

Question Number 43759    Answers: 2   Comments: 0

Question Number 43757    Answers: 0   Comments: 0

Probably if x^n =Am ((a/b)π), x=e^(((2k+a)/(bn))iπ) about 0<(k∈N∪{0})<(n∈N) and b≠0. p.s. Am (0°)=1, Am (90°)=i etc., and s°=(π/(180))s rad(ians)=(π/(180))s.

$$\mathrm{Probably}\:\mathrm{if}\:{x}^{{n}} ={Am}\:\left(\frac{{a}}{{b}}\pi\right),\:{x}={e}^{\frac{\mathrm{2}{k}+{a}}{{bn}}{i}\pi} \\ $$$$\mathrm{about}\:\mathrm{0}<\left({k}\in\mathbb{N}\cup\left\{\mathrm{0}\right\}\right)<\left({n}\in\mathbb{N}\right)\:\mathrm{and}\:{b}\neq\mathrm{0}. \\ $$$$\mathrm{p}.\mathrm{s}.\:{Am}\:\left(\mathrm{0}°\right)=\mathrm{1},\:{Am}\:\left(\mathrm{90}°\right)={i}\:\mathrm{etc}., \\ $$$$\mathrm{and}\:{s}°=\frac{\pi}{\mathrm{180}}{s}\:\mathrm{rad}\left(\mathrm{ians}\right)=\frac{\pi}{\mathrm{180}}{s}. \\ $$

Question Number 43756    Answers: 1   Comments: 0

x^3 +px+q = 0 If equation has all its roots real, find them.

$$\:\:\boldsymbol{{x}}^{\mathrm{3}} +\boldsymbol{{px}}+\boldsymbol{{q}}\:=\:\mathrm{0} \\ $$$$\boldsymbol{{If}}\:\boldsymbol{{equation}}\:\boldsymbol{{has}}\:\boldsymbol{{all}}\:\boldsymbol{{its}}\:\boldsymbol{{roots}} \\ $$$$\boldsymbol{{real}},\:\boldsymbol{{find}}\:\boldsymbol{{them}}. \\ $$

Question Number 43754    Answers: 0   Comments: 0

Question Number 43753    Answers: 0   Comments: 0

Question Number 43749    Answers: 1   Comments: 0

Question Number 43748    Answers: 1   Comments: 0

Question Number 43731    Answers: 1   Comments: 0

Question Number 43713    Answers: 0   Comments: 2

Question Number 43716    Answers: 1   Comments: 3

Question Number 43707    Answers: 1   Comments: 3

Simplify: (x + y + z)(x^(−1) + y^(−1) + z^(−1) ) = (x^(−1) y^(−1) z^(−1) )(x + y)(y + z)(z + x)

$$\mathrm{Simplify}:\:\:\: \\ $$$$\left(\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\right)\left(\mathrm{x}^{−\mathrm{1}} \:+\:\mathrm{y}^{−\mathrm{1}} \:+\:\mathrm{z}^{−\mathrm{1}} \right)\:=\:\left(\mathrm{x}^{−\mathrm{1}} \:\mathrm{y}^{−\mathrm{1}} \:\mathrm{z}^{−\mathrm{1}} \right)\left(\mathrm{x}\:+\:\mathrm{y}\right)\left(\mathrm{y}\:+\:\mathrm{z}\right)\left(\mathrm{z}\:+\:\mathrm{x}\right) \\ $$

Question Number 43706    Answers: 1   Comments: 2

If pqr = 1 Hence evaluate: (1/(1 + e + f^(−1) )) + (1/(1 + f + g^(−1) )) + (1/(1 + g + e^(−1) ))

$$\mathrm{If}\:\:\mathrm{pqr}\:=\:\mathrm{1} \\ $$$$\mathrm{Hence}\:\mathrm{evaluate}:\:\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{e}\:+\:\mathrm{f}^{−\mathrm{1}} }\:\:+\:\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{f}\:+\:\mathrm{g}^{−\mathrm{1}} }\:\:+\:\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{g}\:+\:\mathrm{e}^{−\mathrm{1}} } \\ $$

Question Number 43705    Answers: 0   Comments: 0

Prove that to each quadratic factor in the denominator of the form ax^2 + bx + c which does not have linear factors, there corresponds to a partial fraction of the form ((Ax + B)/(ax^2 + bx + c)) where A and B are constant.

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{to}\:\mathrm{each}\:\mathrm{quadratic}\:\mathrm{factor}\:\mathrm{in}\:\mathrm{the}\:\mathrm{denominator}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form}\: \\ $$$$\mathrm{ax}^{\mathrm{2}} \:+\:\mathrm{bx}\:+\:\mathrm{c}\:\:\:\mathrm{which}\:\mathrm{does}\:\mathrm{not}\:\mathrm{have}\:\mathrm{linear}\:\mathrm{factors},\:\mathrm{there}\:\mathrm{corresponds}\:\mathrm{to} \\ $$$$\mathrm{a}\:\mathrm{partial}\:\mathrm{fraction}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form}\:\:\:\:\frac{\mathrm{Ax}\:+\:\mathrm{B}}{\mathrm{ax}^{\mathrm{2}} \:+\:\mathrm{bx}\:+\:\mathrm{c}}\:\:\:\mathrm{where}\:\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{constant}. \\ $$

Question Number 43702    Answers: 1   Comments: 0

simplify [((12^(1/5) )/(27^(1/5) ))]^(5/2)

$${simplify}\:\:\:\left[\frac{\mathrm{12}^{\mathrm{1}/\mathrm{5}} }{\mathrm{27}^{\mathrm{1}/\mathrm{5}} }\right]^{\mathrm{5}/\mathrm{2}} \\ $$

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