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2 years ago

12

50 points! answer ASAP!! Which of the following is the graph of y = negative (x minus 2) cubed minus 5?

1 answer:

CaHeK987 [17]2 years ago

7 0

Answer: The second image you attached (I also attached the picture of the right answer)

Step-by-step explanation:

In this equation A is negative so you know it is going downward.

the shift is down 5 units and to the right 2.

You can see this shift in the graph I attached

it is also 1 to 1

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Solve the following differential equation: y" + y' = 8x^2

PtichkaEL [24]

Answer:

$y=y_p+y_h = \frac{8}{3}x^3-8x^2+16x+ C_1 + C_2e^{-x}$

Step-by-step explanation:

Let's find a particular solution. We need a function of the form $y= ax^3+bx^2+cx+d$ such that

$y'= 3ax^2+2bx+c$ and

$y''= 6ax+2b$

$y'+y''= 3ax^2+2bx+c+6ax+2b = 3ax^2+x(2b+6a)+(c+2b) = 8x^2$

then, 3a= 8, 2b+6a =0 and c+2b = 0. With the first equation we obtain

a = 8/3 and replacing in the second equation 2b+6(8/3) = 2b + 16 = 0. Then, b = -8. Finally, c = -2(-8) = 16.

So, our particular solution is $y_p= \frac{8}{3}x^3-8x^2+16x$ .

Now, let's find the solution $y_p$ of the homogeneus equation $y''+y'=0$ with the method of constants coefficients. Let $y=e^{\lambda x}$

$y'=\lambda e^{\lambda x}$

$y''=\lambda^2e^{\lambda x}$

then $\lambda e^{\lambda x}+\lambda^2 e^{\lambda x} = 0$

$e^{\lambda x}(\lambda +\lambda^2)= 0$

$(\lambda +\lambda^2)= 0$

$\lambda (1+\lambda)= 0$

$\lambda =0$ and $\lambda)= -1$ .

So, $y_h = C_1 + C_2e^{-x}$ and the solution is

$y=y_p+y_h =\frac{8}{3}x^3-8x^2+16x+ C_1 + C_2e^{-x}$ .

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3 years ago

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Galina-37 [17]

Answer:

The slope is the rise over run, so...

It is 3/1 or just 3.

Step-by-step explanation:

Rise/run is slope

(2,1) is a point on the line, and so is (3,4)

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