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Rasek [7]
3 years ago
13

Please help, thank you.

Mathematics
1 answer:
Gnoma [55]3 years ago
8 0
Lets x = radius 
x^2 + 10^2 = (x+8)^2
x^2 + 100 = x^2 + 16x + 64
16x = 100 - 64
16x = 36
    x = 2.25

answer
B. 2.25
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Step-by-step explanation:

8 0
3 years ago
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What is the value of x? Enter your answer, as a decimal, in the box.<br><br> ___ cm
lora16 [44]

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50.6

Step-by-step explanation:

I'm not sure how to do it, but I took the quiz and it said this was the correct answer


6 0
3 years ago
Solve the following by elimination method <br> 11x+15y+23=0 and 7x-2y-20=0
mr Goodwill [35]
Let 11x + 15y + 23 = 0 be equation (1)
And 7x - 2y - 20 = 0 be equation (2)

Multiply equation (1) by 2:
22x + 30y + 46 = 0

Multiply equation (2) by 15:
105x - 30y - 300 = 0

Add equations (1) and (2):
22x + 105x + 30y - 30y + 46 - 300 = 0
127x - 254 = 0
127x = 254
x = 254/127
[x = 2]

Substitute x = 2 in equation (1) to find y:
11(2) + 15y + 23 = 0
22 + 15y + 23 = 0
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15y = -45
y = -45/15
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Therefore, x = 2 and y = -3.
7 0
3 years ago
Find the general solution of x'1 = 3x1 - x2 + et, x'2 = x1.
Ann [662]
Note that if {x_2}'=x_1, then {x_2}''={x_1}', and so we can collapse the system of ODEs into a linear ODE:

{x_2}''=3{x_2}'-x_2+e^t
{x_2}''-3{x_2}'+x_2=e^t

which is a pretty standard linear ODE with constant coefficients. We have characteristic equation

r^2-3r+1=\left(r-\dfrac{3+\sqrt5}2\right)\left(r+\dfrac{3+\sqrt5}2\right)=0

so that the characteristic solution is

{x_2}_C=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}

Now let's suppose the particular solution is {x_2}_p=ae^t. Then

{x_2}_p={{x_2}_p}'={{x_2}_p}''=ae^t

and so

ae^t-3ae^t+ae^t=-ae^t=e^t\implies a=-1

Thus the general solution for x_2 is

x_2=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}-e^t

and you can find the solution x_1 by simply differentiating x_2.
7 0
3 years ago
This look at pic I need number 22
kupik [55]
I THINK THAT IT IS (A)
3 0
3 years ago
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