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agasfer [191]
3 years ago
15

Simplfy 6x + 7 – 2x + 2 +3x

Mathematics
1 answer:
masha68 [24]3 years ago
6 0

Answer:

<em><u>6</u></em><em><u>x</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>7</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>2</u></em><em><u>x</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>2</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>3</u></em><em><u>x</u></em><em><u> </u></em>

<em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>6</u></em><em><u>x</u></em><em><u> </u></em><em><u>-2x </u></em><em><u>+</u></em><em><u> </u></em><em><u>3</u></em><em><u>x</u></em><em><u> </u></em><em><u>+</u></em><em><u>7</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>2</u></em>

<em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u>9</u></em><em><u>x</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>2</u></em><em><u>x</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>9</u></em>

<em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u>7</u></em><em><u>x</u></em><em><u> </u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>9</u></em><em><u> </u></em>

<em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u>7</u></em><em><u>x</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>-</u></em><em><u>9</u></em>

<em><u> </u></em><em><u> </u></em><em><u>x </u></em><em><u>=</u></em><em><u> </u></em><em><u>-</u></em><em><u>9</u></em><em><u>/</u></em><em><u>7</u></em><em><u> </u></em>

<em><u>hope</u></em><em><u> it</u></em><em><u> helps</u></em>

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Answer:

\frac{y}{x^2}=\sin x+\pi

Step-by-step explanation:

Consider linear differential equation \frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x)

It's solution is of form y\,I.F=\int I.F\,q(x)\,dx where I.F is integrating factor given by I.F=e^{\int p(x)\,dx}.

Given: \frac{1}{x}\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x^2}=x\cos x

We can write this equation as \frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x}=x^2\cos x

On comparing this equation with \frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x), we get p(x)=\frac{-2}{x}\,\,,\,\,q(x)=x^2\cos x

I.F = e^{\int p(x)\,dx}=e^{\int \frac{-2}{x}\,dx}=e^{-2\ln x}=e^{\ln x^{-2}}=\frac{1}{x^2}      { formula used: \ln a^b=b\ln a }

we get solution as follows:

\frac{y}{x^2}=\int \frac{1}{x^2}x^2\cos x\,dx\\\frac{y}{x^2}=\int \cos x\,dx\\\\\frac{y}{x^2}=\sin x+C

{ formula used: \int \cos x\,dx=\sin x }

Applying condition:y(\pi)=\pi^2

\frac{y}{x^2}=\sin x+C\\\frac{\pi^2}{\pi}=\sin\pi+C\\\pi=C

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\frac{y}{x^2}=\sin x+\pi

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So we need to solve:

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3ˣ⁻¹ = 3²ˣ⁺⁴

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