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Question Number 73840 Answers: 0 Comments: 0

Question Number 73838 Answers: 3 Comments: 9

Question Number 73832 Answers: 2 Comments: 7

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Question Number 73817 Answers: 1 Comments: 0

find the solutions of the equation in 0 ≤ θ ≤ π sin2θ = secθ

$${find}\:{the}\:{solutions}\:{of}\:{the}\:{equation} \\ $$$${in}\:\:\mathrm{0}\:\leqslant\:\theta\:\leqslant\:\pi \\ $$$$\:\:{sin}\mathrm{2}\theta\:=\:{sec}\theta \\ $$

Question Number 73816 Answers: 0 Comments: 8

Question Number 74339 Answers: 0 Comments: 2

x^3 +ax^2 +bx+c=0 Let x=((pt+q)/(t+1)) p^3 t^3 +3p^2 qt^2 +3pq^2 t+q^3 +a(t+1)(p^2 t^2 +2pqt+q^2 ) +b(pt+q)(t^2 +2t+1) +c(t^3 +3t^2 +3t+1) = 0 ⇒ (p^3 +ap^2 +bp+c)t^3 +(3p^2 q+ap^2 +2apq+bq+2bp+3c)t^2 +(3q^2 p+aq^2 +2apq+bp+2bq+3c)t +(q^3 +aq^2 +bq+c) = 0 Let coeffs. of t^2 and t be zero. Subtracting and adding them 3pq+a(p+q)+b=0 & 3pq(p+q)+a{(p+q)^2 −2pq} +4apq+3b(p+q)+6c = 0 lets call pq=m , p+q=s ⇒ 3m+as+b=0 ....(i) 3ms+a(s^2 −2m)+4am+3bs+6c=0 ⇒ am+bs+3c=0 ....(ii) ⇒ s=((9c−ab)/(a^2 −3b)) ; m=((b^2 −3ac)/(a^2 −3b)) Now p,q are roots of eq. z^2 −sz+m=0 p,q = (s/2)±(√((s^2 /4)−m)) t^3 =−(((q^3 +aq^2 +bq+c)/(p^3 +ap^2 +bq+c))) (t≠−1) x=((pt+q)/(t+1)) .

$${x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${Let}\:\:\:{x}=\frac{{pt}+{q}}{{t}+\mathrm{1}} \\ $$$${p}^{\mathrm{3}} {t}^{\mathrm{3}} +\mathrm{3}{p}^{\mathrm{2}} {qt}^{\mathrm{2}} +\mathrm{3}{pq}^{\mathrm{2}} {t}+{q}^{\mathrm{3}} \\ $$$$+{a}\left({t}+\mathrm{1}\right)\left({p}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{2}{pqt}+{q}^{\mathrm{2}} \right) \\ $$$$+{b}\left({pt}+{q}\right)\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{1}\right) \\ $$$$+{c}\left({t}^{\mathrm{3}} +\mathrm{3}{t}^{\mathrm{2}} +\mathrm{3}{t}+\mathrm{1}\right)\:\:\:\:=\:\:\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\:\:\left({p}^{\mathrm{3}} +{ap}^{\mathrm{2}} +{bp}+{c}\right){t}^{\mathrm{3}} \\ $$$$+\left(\mathrm{3}{p}^{\mathrm{2}} {q}+{ap}^{\mathrm{2}} +\mathrm{2}{apq}+{bq}+\mathrm{2}{bp}+\mathrm{3}{c}\right){t}^{\mathrm{2}} \\ $$$$+\left(\mathrm{3}{q}^{\mathrm{2}} {p}+{aq}^{\mathrm{2}} +\mathrm{2}{apq}+{bp}+\mathrm{2}{bq}+\mathrm{3}{c}\right){t} \\ $$$$+\left({q}^{\mathrm{3}} +{aq}^{\mathrm{2}} +{bq}+{c}\right)\:=\:\mathrm{0} \\ $$$${Let}\:{coeffs}.\:{of}\:{t}^{\mathrm{2}} \:{and}\:{t}\:{be}\:{zero}. \\ $$$${Subtracting}\:{and}\:{adding}\:{them} \\ $$$$\:\mathrm{3}{pq}+{a}\left({p}+{q}\right)+{b}=\mathrm{0}\:\:\:\& \\ $$$$\mathrm{3}{pq}\left({p}+{q}\right)+{a}\left\{\left({p}+{q}\right)^{\mathrm{2}} −\mathrm{2}{pq}\right\} \\ $$$$\:+\mathrm{4}{apq}+\mathrm{3}{b}\left({p}+{q}\right)+\mathrm{6}{c}\:=\:\mathrm{0} \\ $$$${lets}\:{call}\:\:{pq}={m}\:,\:\:{p}+{q}={s}\:\:\Rightarrow \\ $$$$\:\:\mathrm{3}{m}+{as}+{b}=\mathrm{0}\:\:\:\:....\left({i}\right) \\ $$$$\mathrm{3}{ms}+{a}\left({s}^{\mathrm{2}} −\mathrm{2}{m}\right)+\mathrm{4}{am}+\mathrm{3}{bs}+\mathrm{6}{c}=\mathrm{0} \\ $$$$\Rightarrow\:{am}+{bs}+\mathrm{3}{c}=\mathrm{0}\:\:\:\:....\left({ii}\right) \\ $$$$\Rightarrow\:\:{s}=\frac{\mathrm{9}{c}−{ab}}{{a}^{\mathrm{2}} −\mathrm{3}{b}}\:\:\:;\:\:{m}=\frac{{b}^{\mathrm{2}} −\mathrm{3}{ac}}{{a}^{\mathrm{2}} −\mathrm{3}{b}} \\ $$$$\:{Now}\:\:{p},{q}\:\:{are}\:{roots}\:{of}\:{eq}. \\ $$$$\:\:\:{z}^{\mathrm{2}} −{sz}+{m}=\mathrm{0} \\ $$$$\:\:\:{p},{q}\:=\:\frac{{s}}{\mathrm{2}}\pm\sqrt{\frac{{s}^{\mathrm{2}} }{\mathrm{4}}−{m}} \\ $$$$\:\:{t}^{\mathrm{3}} =−\left(\frac{{q}^{\mathrm{3}} +{aq}^{\mathrm{2}} +{bq}+{c}}{{p}^{\mathrm{3}} +{ap}^{\mathrm{2}} +{bq}+{c}}\right)\:\:\:\:\:\left({t}\neq−\mathrm{1}\right) \\ $$$$\:\:\:{x}=\frac{{pt}+{q}}{{t}+\mathrm{1}}\:. \\ $$

Question Number 74338 Answers: 0 Comments: 0

∫e^(2t) sin e^t dt

$$\int{e}^{\mathrm{2}{t}} \mathrm{sin}\:{e}^{{t}} {dt} \\ $$

Question Number 74337 Answers: 1 Comments: 0

find the contracted form of: ((n),(p) )+2 ((( n)),((p+1)) )+ ((( n)),((p+2)) )

$${find}\:{the}\:{contracted}\:{form}\:{of}: \\ $$$$\begin{pmatrix}{{n}}\\{{p}}\end{pmatrix}+\mathrm{2}\begin{pmatrix}{\:\:\:{n}}\\{{p}+\mathrm{1}}\end{pmatrix}+\begin{pmatrix}{\:\:\:{n}}\\{{p}+\mathrm{2}}\end{pmatrix} \\ $$

Question Number 73805 Answers: 0 Comments: 3

topic binomial theorem Evaluate (2x−6y)^(−8)

$${topic}\:{binomial}\:{theorem} \\ $$$${Evaluate} \\ $$$$\left(\mathrm{2}{x}−\mathrm{6}{y}\right)^{−\mathrm{8}} \\ $$

Question Number 73804 Answers: 0 Comments: 0

Question Number 73800 Answers: 0 Comments: 8

Please draw the shape and find the angles QR = 6 cm RS = 7 cm PS = 4 cm

$$\mathrm{Please}\:\mathrm{draw}\:\mathrm{the}\:\mathrm{shape}\:\mathrm{and}\:\mathrm{find}\:\mathrm{the}\:\mathrm{angles} \\ $$$$\mathrm{QR}\:\:\:=\:\:\mathrm{6}\:\:\mathrm{cm} \\ $$$$\mathrm{RS}\:\:\:=\:\:\mathrm{7}\:\mathrm{cm} \\ $$$$\mathrm{PS}\:\:=\:\:\mathrm{4}\:\mathrm{cm} \\ $$

Question Number 73789 Answers: 3 Comments: 4

Question Number 73787 Answers: 0 Comments: 2

sin 50 + sin 40= ? without tables or calculators

$${sin}\:\mathrm{50}\:+\:{sin}\:\mathrm{40}=\:?\:{without}\:{tables}\:{or}\:{calculators} \\ $$

Question Number 73782 Answers: 1 Comments: 2

Question Number 73775 Answers: 0 Comments: 0

Question Number 73774 Answers: 0 Comments: 0

hello ,show that Σ_(n≥1) (((−1)^n nsin(n))/(1+n^2 ))=((πe^1 −πe^(−1) )/(−2e^π +2e^(−π) )) indication ,Residus Theorem let f(z)=((zsin(z))/((1+z^2 )sin(πz))) have a very nice day!

$$\mathrm{hello}\:,\mathrm{show}\:\mathrm{that} \\ $$$$\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{nsin}\left(\mathrm{n}\right)}{\mathrm{1}+\mathrm{n}^{\mathrm{2}} }=\frac{\pi\mathrm{e}^{\mathrm{1}} −\pi\mathrm{e}^{−\mathrm{1}} }{−\mathrm{2e}^{\pi} +\mathrm{2e}^{−\pi} } \\ $$$$\mathrm{indication}\:,\mathrm{Residus}\:\mathrm{Theorem}\:\mathrm{let} \\ $$$$\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{zsin}\left(\mathrm{z}\right)}{\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)\mathrm{sin}\left(\pi\mathrm{z}\right)} \\ $$$$\mathrm{have}\:\mathrm{a}\:\mathrm{very}\:\mathrm{nice}\:\mathrm{day}! \\ $$

Question Number 73766 Answers: 3 Comments: 0

{ (((1/(x−1))=(2/(y−2))=(3/(z−3)))),((x+2y+3z=56)) :} please help me to solve it in R^3

$$\begin{cases}{\frac{\mathrm{1}}{\mathrm{x}−\mathrm{1}}=\frac{\mathrm{2}}{\mathrm{y}−\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{z}−\mathrm{3}}}\\{\mathrm{x}+\mathrm{2y}+\mathrm{3z}=\mathrm{56}}\end{cases} \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{in}\:\mathbb{R}^{\mathrm{3}} \\ $$

Question Number 73757 Answers: 1 Comments: 1

cos 70°+sin 200°=?

$$\mathrm{cos}\:\mathrm{70}°+\mathrm{sin}\:\mathrm{200}°=? \\ $$

Question Number 73755 Answers: 1 Comments: 0

show that for all integer n , n+1 divides (((2n)),(n) )

$${show}\:{that}\:\:\:{for}\:{all}\:{integer}\:\:{n}\:,\:\:{n}+\mathrm{1}\:{divides}\:\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix} \\ $$

Question Number 73751 Answers: 1 Comments: 1

Find out the value of J=∫_0 ^∞ ∫_0 ^1 (2e^(−2xy) −e^(−xy) )dxdy

$${Find}\:\:{out}\:{the}\:{value}\:{of}\:\:\: \\ $$$$\:\:{J}=\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{2}{e}^{−\mathrm{2}{xy}} −{e}^{−{xy}} \right){dxdy}\: \\ $$

Question Number 73746 Answers: 1 Comments: 1

total numper of words formed by 2 vowels and 3 consonants take from vowels and 5 consonants is equal to ? pleas sir help me ?

$${total}\:{numper}\:{of}\:{words}\:{formed}\:{by}\:\mathrm{2}\:{vowels}\:{and}\:\mathrm{3}\:{consonants}\:{take}\:{from}\:{vowels}\:{and}\:\mathrm{5}\:{consonants}\:{is}\:{equal}\:{to}\:? \\ $$$${pleas}\:{sir}\:{help}\:{me}\:? \\ $$

Question Number 73745 Answers: 0 Comments: 0

total numper of words formed by 2 vowels and 3 consonants take from vowels and 5 consonants is equal to ? pleas sir help me ?

Question Number 73744 Answers: 0 Comments: 0

total numper of words formed by 2 vowels and 3 consonants take from vowels and 5 consonants is equal to ? pleas sir help me ?

Question Number 73737 Answers: 2 Comments: 1

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