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Evgen [1.6K]
2 years ago
14

A way to remember corresponding angles is to remember: *

Mathematics
1 answer:
Montano1993 [528]2 years ago
3 0
Haha ahaha hah ahaha haha
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PLEASE HELP!!<br> A<br> B<br> C<br> D
topjm [15]

Answer:A

Step-by-step explanation:

Its A

5 0
3 years ago
Hey guys, I need help with these questions. I need them done by today! Please, help :( There are 4 questions or more I think, I
Orlov [11]

Answer:

Step-by-step explanation:

Domain x^2 - 9 {Solution: - infinity < x < infinity}

Interval notation (- infinity, infinity)

Range of x^2 - 9 (Solution: f(x) is greater than or equal to - 9)

Interval notation (-9, infinity)

Axis interception points of x^2 - 9:

X- intercepts (3, 0) (-3, 0)

Y-intercepts (0, -9)

Vertex of x^2 - 9: Minimum (0, -9)

Solve for f:

f (x) = x^2 - 9

Step 1: Divide both sides by x.

fx / x = x^2 - 9 / x

f = x^2 - 9 / x

Answer:

f = x^2 - 9 / x

8 0
1 year ago
8n-12=6n+6 I need help figuring out this answer I been having some troble
Gennadij [26K]
Solve the equation with the following steps:
8n-12=6n+6
Subtract 6n from both sides
2n-12=6
Add 12 to both sides
2n=18
Divide both sides by 2
n=9
8 0
3 years ago
A landscaper has enough cement to make a patio with an area of 150 Sq ft. The homeowner wants the length of the patio to be 6 ft
Natasha_Volkova [10]
A=lw=150\\l=w+6\\(w+6)w=150\\w^2+6w=150\\w^2+6w-150=0\\w=\frac{-b+-\sqrt{b^2-4ac}}{2a}\\w=\frac{-(6)+-\sqrt{(6)^2-4(1)(-150)}}{2(1)}\\w=\frac{-6+-\sqrt{36+600}}{2}\\w=\frac{-6+-\sqrt{636}}{2}\\w=\frac{-6+-2\sqrt{159}}{2}\\w=-3+-\sqrt{159}>0\\w=-3+\sqrt{159}\\\\l=(-3+\sqrt{159})+6\\l=3+\sqrt{159}

Width = -3 + sqrt(159)
Length = 3 + sqrt(159)
4 0
3 years ago
Help please!!!!!!!!!!!!!!!!!!!!!!!!11111
yaroslaw [1]
1/3

Hope this helps! :)

(I’m actually doing IXL too lol)
8 0
3 years ago
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