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2 years ago

5

Can someone help me with this?

1 answer:

Natali5045456 [20]2 years ago

8 0

Answer:

ᴀ.)

Step-by-step explanation:

Hope you get it right!! :)

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The length of a rectangle is 5 times its width. Write an expression you would use to find the length when given the width (w). W

nadya68 [22]

Let 5w equal the length and w equal the width.

5w + w = Area
6w = Area

5(20) = Length
Length = 100 inches

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2 years ago

Katena32 [7]

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2 years ago

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

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

5 0

2 years ago

Help please i am toooo smart to know

zalisa [80]

Answer:

10.6 repeating

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Kiran walked at a constant speed. He walked 1 mile in 15 minutes. Which of these equations represents the distance d (in miles)

galben [10]

Answer: $d=\dfrac{t}{15}$ , where d= distance that Kiran walks in t minutes.

Step-by-step explanation:

Given: Kiran walked at a constant speed.

He walked 1 mile in 15 minutes.

The, distance walked by Kiran in 1 minute = $\dfrac1{15}$ mile

The distance that Kiran walks in t minutes = $t\times\dfrac{1}{15}$ miles

If d= distance that Kiran walks in t minutes

The required equation: $d=\dfrac{t}{15}$

5 0

2 years ago