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garik1379 [7]
3 years ago
15

180 is the lcm of 12 and a number n . What are the possible values of n?

Mathematics
1 answer:
love history [14]3 years ago
6 0
12|\fbox2\\.\ 6|\fbox2\\.\ 3|\fbox3\\.\ 1|\\----------\\180|\fbox2\\.\ 90|\fbox2\\.\ 45|\fbox3\\.\ 15|3\\.\ \ 5|5\\.\ \ 1|\\\\Answer:\\5\times3\times3=45\\5\times3\times3\times2=90\\5\times3\times3\times2\times2=180
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Lilly and Maggie decorated their windows with twinkle lights. Maggie used 51 more twinkle lights than Lilly. Fill in the box wit
nadya68 [22]
We know that
m-------------------- > <span>the amount of twinkle lights Maggie used
l--------------------- >  </span><span>the amount of twinkle lights Lilly used
l=70
m=51+l
therefore
m=51+70=121
m=121

the answer is 121 (</span>amount of twinkle lights Maggie used)
6 0
4 years ago
Solution for dy/dx+xsin 2 y=x^3 cos^2y
vichka [17]
Rearrange the ODE as

\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y
\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3

Take u=\tan y, so that \dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}.

Supposing that |y|, we have \tan^{-1}u=y, from which it follows that

\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}
\sec^2y=1+\tan^2y=1+u^2

So we can write the ODE as

\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3

which is linear in u. Multiplying both sides by e^{x^2}, we have

e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}
\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}

Integrate both sides with respect to x:

\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx
e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx

Substitute t=x^2, so that \mathrm dt=2x\,\mathrm dx. Then

\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt

Integrate the right hand side by parts using

f=t\implies\mathrm df=\mathrm dt
\mathrm dg=e^t\,\mathrm dt\implies g=e^t
\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)

You should end up with

e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C
u=\dfrac{x^2-1}2+Ce^{-x^2}
\tan y=\dfrac{x^2-1}2+Ce^{-x^2}

and provided that we restrict |y|, we can write

y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)
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John makes $400 per week. What does it mean for him to get a 150% salary raise? Explain in words and then give an example showin
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He’s gotten a 200$ raise
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Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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Answer:

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