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Citrus2011 [14]

3 years ago

11

Can you help me find the measure of angle B plss

1 answer:

wariber [46]3 years ago

4 0

Angie B is 128 degrees. Because this is a supplementary angle you can find the unknown value by subtracting 52 from 180

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Cisco wants to purchase three items at the sporting goods store. The items he wants to buy are football pants for $21.99, footba

Sergeeva-Olga [200]

Answer: $\$73.43$

Step-by-step explanation:

According to the information provided in the exercise, the prices of each item he wants to purchase in the sporting goods store are:

$Football\ pants=\$21.99\\\\ Football\ pads=\$25.95\\\\Football\ cleats=\$25.49$

Therefore you need to add these prices to solve this exercise. The sum will be the total cost for these three items.

You get that the sum is:

$Total\ cost=\$21.99+\$25.95+\$25.49\\\\Total\ cost=\$73.43$

Then, the three items that Cisco wants to purchase will cost $73.43.

8 0

4 years ago

Solve: <br> 10c + 20 = 40

Natali5045456 [20]

Subtract 20 from 40 then divide by 10

C=2

8 0

3 years ago

Read 2 more answers

1.7 meters to millimeters_____m =

Aloiza [94]

1700m is the correct answer

7 0

3 years ago

Read 2 more answers

Use the method of undetermined coefficients to find the general solution to the de y′′−3y′ 2y=ex e2x e−x

djverab [1.8K]

I'll assume the ODE is

$y'' - 3y' + 2y = e^x + e^{2x} + e^{-x}$

Solve the homogeneous ODE,

$y'' - 3y' + 2y = 0$

The characteristic equation

$r^2 - 3r + 2 = (r - 1) (r - 2) = 0$

has roots at $r=1$ and $r=2$ . Then the characteristic solution is

$y = C_1 e^x + C_2 e^{2x}$

For nonhomogeneous ODE (1),

$y'' - 3y' + 2y = e^x$

consider the ansatz particular solution

$y = axe^x \implies y' = a(x+1) e^x \implies y'' = a(x+2) e^x$

Substituting this into (1) gives

$a(x+2) e^x - 3 a (x+1) e^x + 2ax e^x = e^x \implies a = -1$

For the nonhomogeneous ODE (2),

$y'' - 3y' + 2y = e^{2x}$

take the ansatz

$y = bxe^{2x} \implies y' = b(2x+1) e^{2x} \implies y'' = b(4x+4) e^{2x}$

Substitute (2) into the ODE to get

$b(4x+4) e^{2x} - 3b(2x+1)e^{2x} + 2bxe^{2x} = e^{2x} \implies b=1$

Lastly, for the nonhomogeneous ODE (3)

$y'' - 3y' + 2y = e^{-x}$

take the ansatz

$y = ce^{-x} \implies y' = -ce^{-x} \implies y'' = ce^{-x}$

and solve for $c$ .

$ce^{-x} + 3ce^{-x} + 2ce^{-x} = e^{-x} \implies c = \dfrac16$

Then the general solution to the ODE is

$\boxed{y = C_1 e^x + C_2 e^{2x} - xe^x + xe^{2x} + \dfrac16 e^{-x}}$

6 0

1 year ago

How much lower was the 6am temperature in Juneau than in Los Angeles

adell [148]

A lot like shoot i think 20 or more

6 0

3 years ago