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Lubov Fominskaja [6]
3 years ago
8

HELPPPPP PLSS- What is the area of this figure? 9 yd 5 yd 7 yd 3 yd 2 yd 6 yd square yards​

Mathematics
1 answer:
Cerrena [4.2K]3 years ago
5 0

Answer:

99

Step-by-step explanation:

9x5 + 7X6 + 3X4 + 2X6

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R - 15 = 77<br> I need to show my work
Dmitrij [34]

Answer:

R - 15 = 77

R = 77 + 15

R = 92

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3 years ago
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Which expression have the same value as the product of 2.7 and 4
Maurinko [17]

Answer:

Step-by-step explanation:

2.7*4=10.80

27*0.4=10.80 we multiplied 27 by 10 but divide 4 by 10

0.27*40=10.80

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Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
irakobra [83]

Answer:

a-9=20

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6 0
3 years ago
Alena had 4 packs of gum with 6 pieces of gum in each pack. Hugh had 3 packs of gum with 5 pieces in each. Alena and Hugh have 1
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8 0
3 years ago
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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
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