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goldenfox [79]
3 years ago
15

Use substitution to solve the system of equations :

Mathematics
1 answer:
mote1985 [20]3 years ago
3 0

Answer:

x=1

y=-4

Step-by-step explanation:

4x+y=0  ->    y= -4x

2(-4x)+x=-7

-8x+  x=-7

-7x=-7

x=-7/-7=1

y=-4 . 1 = -4

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X - (-3) is the same as x + (-3).<br><br><br> True or False
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False

Step-by-step explanation:

x - (-3) = x + 3

x + (-3) = x - 3

8 0
3 years ago
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you drop a ball that bounces back up to 95 in. after the first bounce. after each bounce, the ball reaches a height that is 75%
k0ka [10]

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187

Step-by-step explanation:

7 0
3 years ago
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How would I do this
xz_007 [3.2K]
When,
correct is 0 then p is 0/20.
correct is 1 p is 1/20
correct is 2 p is 2/20 that is 1/10
correct is 3 p is 3/20.
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5 0
4 years ago
Evaluate the integral. (remember to use absolute values where appropriate. Use c for the constant of integration.) 5 cot5(θ) sin
ozzi

I=5\int \frac{cos^{4}\theta }{sin\theta }\times cos\theta d\theta \\\\I=5\int \left ( 1-sin^{2}\theta  \right )^{2}\times \frac{cos\theta }{sin\theta }d\theta \\put\ \sin\theta =t\\\\dt=cos\theta d\theta \\\\I=5\int\frac{t^{4}+1-2t^{2}}{t}dt\ \ \ \ \ \ \ \ \ \ \because (a-b)^2=a^2+b^2-2ab\\\\I=5\left ( \int t^{3}dt + \int \frac{1}{t} -2\int t \right )dt

by using the integration formula

we get,

\\I=5\left ( \frac{t^{4}}{4} +logt -t^{2}\right )\\\\I=\frac{5}{4}t^{4}+5\log t-5t^{2}+c

now put the value of t=\sin\theta in the above equation

we get,

\int 5\cot^5\theta \sin^4\theta d\theta=\frac{5}{4}sin^{4}\theta+5\log \sin\theta - 5sin^{2} \theta+c

hence proved

7 0
3 years ago
Guys Pls help <br><br><br><br><br><br> 2+2=??????????????????/
Mashcka [7]
Dang this is so hard bro but I think it’s 4
5 0
3 years ago
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