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3 years ago

7

Jolene bought a savings certificate that earned 2% in simple interest per year. She earned a total of $240 in interest over four

years. How much was the certificate worth, with interest, after four years?

1 answer:

vovikov84 [41]3 years ago

3 0

Answer:

dxcfghjkm,

Step-by-step explanation:

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Which quantity is proportional to 21/3

BARSIC [14]

$\frac{7}{1}$
You can simplify the top and bottom by 3

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4 years ago

Ryan has his own business selling watches, and he wants to monitor his profit per watch sold. The expression <img src="https://t

Umnica [9.8K]

Answer:

(-x² + 99x +200)/x

Step-by-step explanation:

Profit per watch = (200 - x)/x

Profit per band = 100 – x

New profit = (200 - x)/x + 100 – x

New profit = (200 - x)/x + x(100 – x)/x

New profit = (200 - x + 100x – x²)/x

New profit = (-x² + 99x +200)/x

6 0

3 years ago

Read 2 more answers

(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor <em>µ(x, y)</em> such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let <em>µ(x, y)</em> = <em>µ(x)</em> be independent of <em>y</em>, then <em>∂µ/∂y</em> = 0 and we can solve for <em>µ</em> :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$

The modified DE,

$\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0$

is now exact:

$\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}$

So we look for a solution of the form <em>F(x, y)</em> = <em>C</em>. This solution is such that

$\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}$

Integrate both sides of the first condition with respect to <em>x</em> :

$F(x,y) = -e^{-x}y - \dfrac1x + g(y)$

Differentiate both sides of this with respect to <em>y</em> :

$\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C$

Then the general solution to the DE is

$F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}$

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3 years ago

Crystal is 12 years old and Alexa is 6 years old. If crystal runs 100 yards how far does Alexa run? How far do they run altogeth

Nata [24]

I think its 50 because they are half of each others ages so half of 100 is 50 hope it helps sorry if im wrong

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3 years ago

(2x^2y+14xy)-(-8x^2y-9xy+5)

lapo4ka [179]

The answer is 10x^2y + 23xy - 5

8 0

3 years ago

Read 2 more answers