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Makovka662 [10]
3 years ago
11

20 points and will give brainliest!!

Mathematics
1 answer:
FromTheMoon [43]3 years ago
4 0

Answer:

C=0.65M+22.65

Step-by-step explanation:

The standard charge includes an initial fee and plus an addditional fee for each mile driven and is given by:

S=0.40M+17.75

The insurance charge is given by:

I=0.25M+4.90

So, the total charge C will be the sum of the standard charge S and the insurance charge I:

C=S+I

Substitute:

C=(0.40M+17.75)+(0.25M+4.90)

Rearrange:

C=(0.40M+0.25M)+(17.75+4.90)

Add:

C=0.65M+22.65

And we have acauired our equation relating C to M!

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(a)

\displaystyle\int_3^\infty \frac{\mathrm dx}{x^2+9}=\lim_{b\to\infty}\int_{x=3}^{x=b}\frac{\mathrm dx}{x^2+9}

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\displaystyle\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\frac{3\sec^2(t)}{(3\tan(t))^2+9}\,\mathrm dt=\frac13\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\mathrm dt

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\displaystyle \sum_{n=3}^\infty \frac1{n^2+9}

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\displaystyle\sum_{n=3}^\infty\frac1{n^2}

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\displaystyle \sum_{n=1}^\infty \frac{(-1)^n (n^2+9)}{e^n}

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\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n (n^2+9)}{e^n}\right|=\sum_{n=1}^\infty \frac{n^2+9}{e^n} < \sum_{n=1}^\infty \frac{n^2}{e^n} < \sum_{n=1}^\infty \frac1{e^n}=\frac1{e-1}

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