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klasskru [66]
3 years ago
6

Your salary is $25,000 you just received a 2.5% pay increase for the year. What is your new salary?

Mathematics
1 answer:
seraphim [82]3 years ago
8 0
Hey there!

Here is your answer:

<u><em>The proper answer to this question is "$</em></u><span><em /><u><em>25,625".</em></u>
</span>
Reason:

<u><em>In order to add a 2.5% increase on a pay check you have to multiply 25,000 by 1.025.....</em></u>

<u><em>25,000×1.025=$</em></u><span><u><em>25,625</em></u>

Therefore my new salary is $</span><span>25,625!

If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit</span>
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Answer:

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6 0
3 years ago
Use Green's Theorem to evaluate the line integral along the given positively oriented curve. ∫C (3y +5e√x)dx + (10x + 3 cos
elena-s [515]

By Green's theorem, the line integral

\displaystyle \int_C f(x,y)\,\mathrm dx + g(x,y)\,\mathrm dy

is equivalent to the double integral

\displaystyle \iint_D \frac{\partial g}{\partial x} - \frac{\partial f}{\partial y} \,\mathrm dx\,\mathrm dy

where <em>D</em> is the region bounded by the curve <em>C</em>, provided that this integrand has no singularities anywhere within <em>D</em> or on its boundary.

It's a bit difficult to make out what your integral should say, but I'd hazard a guess of

\displaystyle \int_C \left(3y+5e^{-x}\right)\,\mathrm dx + \left(10x+3\cos\left(y^2\right)\right)\,\mathrm dy

Then the region <em>D</em> is

<em>D</em> = {(<em>x</em>, <em>y</em>) : 0 ≤ <em>x</em> ≤ 1 and <em>x</em> ² ≤ <em>y</em> ≤ √<em>x</em>}

so the line integral is equal to

\displaystyle \int_0^1\int_{x^2}^{\sqrt x} \frac{\partial\left(10x+3\cos\left(y^2\right)\right)}{\partial x} - \frac{\partial\left(3y+5e^{-x}\right)}{\partial y}\,\mathrm dy\,\mathrm dx \\\\ = \int_0^1 \int_{x^2}^{\sqrt x} (10-3)\,\mathrm dy\,\mathrm dx \\\\ = 7\int_0^1 \int_{x^2}^{\sqrt x} \mathrm dy\,\mathrm dx

which in this case is 7 times the area of <em>D</em>.

The remaining integral is trivial:

\displaystyle 7\int_0^1\int_{x^2}^{\sqrt x}\mathrm dy\,\mathrm dx = 7\int_0^1y\bigg|_{y=x^2}^{y=\sqrt x}\,\mathrm dx \\\\ = 7 \int_0^1\left(\sqrt x-x^2\right)\,\mathrm dx \\\\ = 7 \left(\frac23x^{3/2}-\frac13x^3\right)\bigg|_{x=0}^{x=1} = 7\left(\frac23-\frac13\right) = \boxed{\frac73}

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3 years ago
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Answer:

What is the average rate of change for this exponential function for the interval from x=0 to x=2

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m = (4 - 1) / (2 - 0)

m = 3 / 2

m = 1.5

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Answer:

it equals out to 0.14285714285 or in fraction form is 1⁄7

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