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Question Number 113503 Answers: 0 Comments: 1

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Question Number 113488 Answers: 2 Comments: 0

prove that ((sin(2a)cos(2a))/(cos(4a))) = (1/2) tan(4a)

$${prove}\:{that} \\ $$$$\frac{{sin}\left(\mathrm{2}{a}\right){cos}\left(\mathrm{2}{a}\right)}{{cos}\left(\mathrm{4}{a}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:{tan}\left(\mathrm{4}{a}\right) \\ $$

Question Number 113486 Answers: 1 Comments: 1

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Question Number 113465 Answers: 0 Comments: 7

Question Number 113464 Answers: 1 Comments: 0

(x^2 (d^2 y/dx^2 ) +x (dy/dx) + 1).y = 0

$$\:\left({x}^{\mathrm{2}} \:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+{x}\:\frac{{dy}}{{dx}}\:+\:\mathrm{1}\right).{y}\:=\:\mathrm{0} \\ $$

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Question Number 113451 Answers: 3 Comments: 0

If 2x=a^n +a^(−n) and 2y=a^n −a^(−n) calculate the value of x^2 −y^(2 ) in its simplest form

$$\mathrm{If}\:\mathrm{2x}=\mathrm{a}^{\mathrm{n}} +\mathrm{a}^{−\mathrm{n}} \:\mathrm{and}\:\mathrm{2y}=\mathrm{a}^{\mathrm{n}} −\mathrm{a}^{−\mathrm{n}} \:\mathrm{calculate} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}\:} \:\mathrm{in}\:\mathrm{its}\:\mathrm{simplest}\:\mathrm{form} \\ $$

Question Number 113444 Answers: 1 Comments: 0

y′′−2ay′+(1+a^2 )y=te^(at) +sint

$$\mathrm{y}''−\mathrm{2ay}'+\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \right)\mathrm{y}=\mathrm{te}^{\mathrm{at}} +\mathrm{sint} \\ $$

Question Number 113438 Answers: 2 Comments: 0

lim_(x→∞) x(5^(1/x) −1) ?

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}\left(\mathrm{5}^{\frac{\mathrm{1}}{{x}}} −\mathrm{1}\right)\:? \\ $$

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Question Number 113418 Answers: 3 Comments: 1

∫ ((3x−2)/( (√(x^2 +2x+26)))) dx

$$\:\int\:\frac{\mathrm{3x}−\mathrm{2}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{26}}}\:\mathrm{dx} \\ $$

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Question Number 113393 Answers: 0 Comments: 6

loving questions of the form “if ... then find the sum/product/etc. of... so please solve these: (1) if γ and λ are the solutions of x^2 +x−12=0 then find coshλ−cotγ (2) if a+b=2 and a−b=0 then find ∫x^((a+b)/(2ab)) ln(−e^(iπa) −x)ln(e^(cos^(−1) b) −x)dx are you intelligent enough? then please please please sir or madam help me!!! I need an answer urgentliest!!!! good luck! (c) by HeR MaJε∫tY 20200913

$${loving}\:{questions}\:{of}\:{the}\:{form} \\ $$$$``{if}\:...\:{then}\:{find}\:{the}\:{sum}/{product}/{etc}.\:{of}... \\ $$$${so}\:{please}\:{solve}\:{these}: \\ $$$$\left(\mathrm{1}\right) \\ $$$${if}\:\gamma\:{and}\:\lambda\:{are}\:{the}\:{solutions}\:{of} \\ $$$${x}^{\mathrm{2}} +{x}−\mathrm{12}=\mathrm{0}\:{then}\:{find}\:{cosh}\lambda−{cot}\gamma \\ $$$$\left(\mathrm{2}\right) \\ $$$${if}\:{a}+{b}=\mathrm{2}\:{and}\:{a}−{b}=\mathrm{0}\:{then}\:{find} \\ $$$$\int{x}^{\frac{{a}+{b}}{\mathrm{2}{ab}}} {ln}\left(−{e}^{{i}\pi{a}} −{x}\right){ln}\left({e}^{{cos}^{−\mathrm{1}} {b}} −{x}\right){dx} \\ $$$${are}\:{you}\:{intelligent}\:{enough}? \\ $$$${then}\:{please}\:{please}\:{please}\:{sir}\:{or}\:{madam} \\ $$$${help}\:{me}!!!\:{I}\:{need}\:{an}\:{answer}\:\boldsymbol{\mathrm{urgentliest}}!!!! \\ $$$${good}\:{luck}! \\ $$$$\left({c}\right)\:{by}\:\mathbb{H}\mathfrak{e}\mathscr{R}\:\mathfrak{M}\boldsymbol{{a}}\mathbb{J}\epsilon\int{t}\mathscr{Y}\:\:\mathrm{20200913} \\ $$

Question Number 113357 Answers: 0 Comments: 7

lim_(x→∞) (ln(ln(lnx))))^(1/x)

$$\left.{lim}_{{x}\rightarrow\infty} \left({ln}\left({ln}\left({lnx}\right)\right)\right)\right)^{\frac{\mathrm{1}}{{x}}} \\ $$

Question Number 113355 Answers: 1 Comments: 1

A rectangular cardboard is 8cm long and 6cm wide. What is the least number of beads you can arrange on the board such that there are at least two of the beads that are less than (√(10))cm apart.

$$\mathrm{A}\:\mathrm{rectangular}\:\mathrm{cardboard}\:\mathrm{is}\:\mathrm{8cm}\:\mathrm{long} \\ $$$$\mathrm{and}\:\mathrm{6cm}\:\mathrm{wide}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{least} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{beads}\:\mathrm{you}\:\mathrm{can}\:\mathrm{arrange}\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{board}\:\mathrm{such}\:\mathrm{that}\:\mathrm{there}\:\mathrm{are}\:\mathrm{at}\:\mathrm{least} \\ $$$$\mathrm{two}\:\mathrm{of}\:\mathrm{the}\:\mathrm{beads}\:\mathrm{that}\:\mathrm{are}\:\mathrm{less}\:\mathrm{than} \\ $$$$\sqrt{\mathrm{10}}\mathrm{cm}\:\mathrm{apart}. \\ $$

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