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daser333 [38]
2 years ago
11

What is the solution of 3x+9<15

Mathematics
1 answer:
Zolol [24]2 years ago
3 0

Answer:

Third one

Step-by-step explanation:

<u>Two </u>things to solve      3x+9 <= 15

                                        3x <= 6

                                         <u> x <= 2</u>

      <u>   and</u>   - (3x+ 9) <= 15

                  - 3x - 9 <= 15

                   - 3x <= 24

                      3x >= -24

                     <u>   x > = - 8 </u>

<u />

<u />

<u>s o   -8 <= x <= 2 </u>

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Which is the correct equation for x:y=8:1<br><br> See picture attached.
statuscvo [17]

Answer:

Step-by-step explanation:

Means of means = means of extremes 8y = x

x = 8y

Option B is the correct answer

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3 years ago
Bill is a car salesman. He earns $200 for every car he sells plus a 3% commission. If Bill sells 3 cars in one week for a total
katovenus [111]
The total he earns is the sum of two parts. One part is the $200 per car. The other part is the 3% commission.

total earnings = $200 per car + 3% commission on amount of sales

Let's look at the $200 per car first.
He sold 3 cars.
He earns $200 for each car he sells, so just from that he earns
3 * $200 = $600.

Now we look at the 3% commission.
He earns 3% of the total amount of sales.
The total amount of sales for the 3 cars was $32,343.
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3% of $32,3243 = 3% * $32,343 = 0.03 * $32,343 = $970.29

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7 0
3 years ago
Read 2 more answers
What algebraic expression is a polynomial with a degree of 4
Monica [59]

Answer:

Any expression that has an X with an exponent of 4 and no other higher exponent.

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(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}

5 0
3 years ago
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