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Zolol [24]
3 years ago
13

How do i find the quotient of rational expressions???

Mathematics
1 answer:
alisha [4.7K]3 years ago
5 0
First, turn it into a multiplication problem by multiplying by the reciprocal of the divisor. Then, you can cancel common factors in the numerator and denominator to make things easier to work with. Finally, multiply<span> to get your final answer!</span>
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What's (7x/2-5x+3)+(2x/2+4x-6)
Anika [276]
I'm assuming that when you wrote "(7x/2-5x+3)+(2x/2+4x-6)," you actually meant "<span>(7x^2-5x+3)+(2x^2+4x-6).  Correct me if I'm wrong here.

</span><span>+(7x^2-5x+3)
</span><span>+(2x/2+4x-6)
-------------------
=9x^2 - x - 3 (answer) </span>
5 0
3 years ago
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TEA [102]
P 14 is r=14 and P 16 is r=12
7 0
3 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)
5 0
3 years ago
Hello, could you help me with my math exercise? thank you in advance
Natali [406]

Answer:

100(x - 1)

Explanation:

Let's begin writing down everything

(X×(-11)+5)×(-9)+X-55

Now we can multiply

-9(-11X+5)+X-55

99X-45+X-55

And so we get

100X-100

Factorize 100

100(x-1)

4 0
3 years ago
Read 2 more answers
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Marat540 [252]

Answer:

lol brainliest

Step-by-step explanation:

8 0
3 years ago
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