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3 years ago

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Can someone help me with this problem

1 answer:

melomori [17]3 years ago

6 0

After 1 year it is 200, 3 years it is 600,5years=1000,193

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Need help with 8-10 didn’t understand this

castortr0y [4]

Answer:

8. F) -5
9. D) (-2,3)
10. J) -10

7 0

3 years ago

Read 2 more answers

Multiply.<br> 2x(x2 + 2x - 6)

Stels [109]

<span>2x(x2 + 2x - 6)
</span>First add what is in the "(" wich is 2x+2x
<span>2x(4x - 6)
</span>Second multiply the 2x by the 4x & -6
<span>8x-12x
Now subtract the -12x from the 8x
8x-12x = -4x

</span>

8 0

3 years ago

Read 2 more answers

What is the average rate of change of the function on the interval from x = 1 to x = 2?

neonofarm [45]

x=1 f = 10(5.5)(1) = 55

x=2, f = 10(5.5)(2) = 110

110-55 = 55

2-1 =1

55/1 = 55

average rate of change = 55

6 0

3 years ago

Find all the solutions for the equation:

Contact [7]

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

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago

SUPREME COURT The first Chief Justice of the U.S. Supreme Court, John Jay, served 2079 days

sashaice [31]

Answer:

John Roberts must serve a minimum of 12, 543 days to surpass John Mrshall's record of justice.

Step-by-step explanation:

John Marshall = 2079 + 10, 463 = 12, 542 days

Now, we have to add 1 more day to pass John Marshall because if we do not, we would only be on the same level as John Marshall and NOT EXCEEDING him.

4 0

3 years ago