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dedylja [7]
3 years ago
7

Solve 6x + 2 = 5x + 17 for x if there is a solution.

Mathematics
1 answer:
labwork [276]3 years ago
3 0

Answer: x=15


Step-by-step explanation:

To solve problems like this you need to simplify the equations on both sides and then isolate your variable x.


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A quarterly telephone bill consist of N$79.15 rental plus 4.7 cents for each dialled unit. Sales tax is added at 15%. What is th
lidiya [134]
$4.7 × 915 = 4300.5
$4300.5 + 79.15 = 4379.65
15% of 4379.65 = 656.9475
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Mrs. Jones' Bill was $5036.63
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8 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}

5 0
3 years ago
Please help me with this. ​
masya89 [10]

Answer:

a) 48 b) 132 c) 180 d) 228 e) 312

Step-by-step explanation:

3 0
2 years ago
Y = 10x + 6 sollve this by doing what
Vladimir79 [104]

Answer:

x = -3/5

Step-by-step explanation:

y = 10x + 6

0 (- 6) = 10x + 6 (- 6)

-6/10 = 10x/10

-6/10 = x

-3/5 = x

6 0
2 years ago
Side BC is 4 meters long. Side CD is twice the length of the side BC. What is the length of side LM. Explain your reasoning.
Misha Larkins [42]

Answer:

LM = 18

Step-by-step explanation:

BC = 4

CD is twice that, so CD is 8

Now, remembering that the area of ABCD = 32 and JKLM = 72, divide 72 by 32.

You get 2.25. This is how much bigger JKLM is than ABCD.

Using that, multiply the length of CD (which is 8) by 2.25 to get the length of LM.

You'll get 18.

Hope this helps!

4 0
2 years ago
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