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3 years ago

7

Please help me one question left

2 answers:

siniylev [52]3 years ago

6 0

Answer: is D

Step-by-step explanation:

madreJ [45]3 years ago

4 0

Answer:

thanks for points

Step-by-step explanation:

too easy boi

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Look at the screenshot please for question!<br> Answer choices:<br> 18<br> 20<br> 21<br> 25

katen-ka-za [31]

Answer:

The Answer is 18 then 25 then 21 then 20

Step-by-step explanation:

5 0

3 years ago

URGENTTT Whos correct and Why?

andreyandreev [35.5K]

Bills right because the y-intercept is 50 so I’m think what he’s saying about the rocket being launched 50 feet above water is right

4 0

3 years ago

How do I solve this math problem?

Nat2105 [25]

Answer: 4x² + 4x - 2 - $\frac{3}{2x+3}$

<u>Step-by-step explanation:</u>

You can use either synthetic or long division. I am using synthetic division:

2x + 3 = 0 ⇒ x = $-\frac{3}{2}$

$-\frac{3}{2}$ | 4 10 4 -6

<u>| ↓ -6 -6 3 </u>

4 4 -2 -3 ← <em>-3 is the remainder</em>

↓ ↓ ↓

4x² +4x -2 ← <em>factored polynomial</em>

3 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

Can you explain why in detail please?

Igoryamba

it would be the graph on the right.

Since the equation ends with +2 the line would be 2 above the x axis

4 0

2 years ago