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3 years ago

8

PLEASE HELP!! will give brainliest. answer choices and question are in the photo.

1 answer:

Ivan3 years ago

4 0

Answer:

-17,-6and 1,-6

Step-by-step explanation:

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Tom saves 50% of his paycheck. If his<br> pay is $148, how much will he save?

Whitepunk [10]

Answer:

triangllllllllllllllllllllllllllllllllllllleeeee

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Given ƒ(x) = x + 10 and g(x) = x2 + 5x - 1, find ƒ(3) + g(5).

SashulF [63]

If I’m not mistaken it should be 8x+25. :)
Hope this helps!

7 0

2 years ago

Diego was trying to write (5^2)^4 with a single exponent and wrote (5^2)^4=5^2+^4=5^6. What should the answer be? Explain Diego'

o-na [289]

Answer:

5^8

Step-by-step explanation:

Diego added the exponents. This was an error. If he was simplifying

5^2 × 5^4, then he could add the exponents and get a correct answer. But his problem had a power raised to a power. In this case, you multiply the exponents to simplify.

(5^2)^4 means

5^2×5^2×5^2×5^2

which is

5×5×5×5×5×5×5×5

which is 5^8.

4 0

2 years ago

I don't know if this is answerable over text-..<br>

andrew11 [14]

Step-by-step explanation:

i wish I could help but no idea

8 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