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VladimirAG [237]
3 years ago
14

DOES ANYONE KNOW HOW TO DO THIS???????? WILL MARK BRAINLISET!!!!!!!!!!!!

Mathematics
1 answer:
skelet666 [1.2K]3 years ago
3 0

Answer:

1. Fig. 4

2. Fig. 4

3. Fig. 1

Step-by-step explanation:

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A(n)_____is a mathematical sentence that compares expressions that are equal
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A swimming pool has an area of 40 square meters. It’s length is 8 meters. What is the width?
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7 0
4 years ago
Read 2 more answers
Y=-x^2+2x+10<br> y=x+2<br><br> Substitution<br> Need ASAP
aniked [119]

Answer:

x_1=\frac{1+\sqrt{33} }{2},y_1=\frac{5+\sqrt{33} }{2}\\ \\ x_2=\frac{1-\sqrt{33} }{2},y_2=\frac{5-\sqrt{33} }{2}

Explanation:

You must find the solution of the system using the substitution method.

System:

y=-x^2+2x+10\\\\ y=x+2

Substitute x+2 for y:

x+2=y=-x^2+2x+10

Pass all the terms to one side:

x^2-2x+x+2-10\\ \\ x^2-x-8=0

Use the quadratic formula to solve:

x=\frac{-b+/-\sqrt{b^2-4ac} }{2a}\\ \\ x=\frac{-(-1)+/-\sqrt{(-1)^2-4(1)(-8)} }{2(1)}\\ \\ x=\frac{1+/-\sqrt{1+32} }{2}

x_1=\frac{1+\sqrt{33} }{2}\\ \\x_2=\frac{1-\sqrt{33} }{2}

Use y = x  + 2 to find the value of y:

y_1=2+x_1=2+\frac{1+\sqrt{33} }{2}\\ \\y_1=\frac{5+\sqrt{33} }{2} \\\\y_2=2+x_2=2+\frac{1-\sqrt{33} }{2}\\ \\ y_2=\frac{5-\sqrt{33} }{2}

6 0
3 years ago
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