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3 years ago

Does anybody know the awnser to -12x+8+5x=36

Mathematics

1 answer:

Nataly [62]3 years ago

5 0

Answer:

x = -4

Step-by-step explanation:

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What are the coordinates of the midpoint of GH with endpoints (-2,5) and H(4,1)?

kozerog [31]

Answer:

(1, 3)

Step-by-step explanation:

Given coordinates of 2 endpoints, G(-2, 5) and H(4, 1), midpoint of AB is calculated as shown below:

Midpoint (M) of GH, for G(-2, 5) and H(4, 1) is given as:

$M(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$

Let $G(-2, 5) = (x_1, y_1)$

$H(4, 1) = (x_2, y_2)$

Thus:

$M(\frac{-2 + 4}{2}, \frac{5 + 1}{2})$

$M(\frac{2}{2}, \frac{6}{2})$

$M(1, 3)$

Coordinates of the midpoint of GH = (1, 3).

6 0

2 years ago

Please answer asap!!!! will mark brainiest

Inga [223]

Answer:

x=2

Step-by-step explanation:

We know the only number that gets us back to 0 after subtracting 4 is 4, because 4-4=0. The square root of 4 is 2. Hope my explanation wasn't too confusing!

2²=4

4-4=0

3 0

3 years ago

Read 2 more answers

Compute the permutations and combinations. 4 P 4 1 4 24

lawyer [7]

The answer is a.
4 P 4 = 24

4 0

3 years ago

If the endpoints of the diameter of a circle are (−8, −6) and (−4, −14), what is the standard form equation of the circle? A) (x

Andrei [34K]

Answer:

$\large\boxed{A.\ (x+6)^2+(y+10)^2=20}$

Step-by-step explanation:

The standard form of an equation of a circle:

$(x-h)^2+(y-k)^2=r^2$

(h, k) - center

r - radius

We have the endpoints of the diameter of a circle (-8, -6) and (-4, -14).

The midpoint of a diameter is a center of a circle.

The formula of a midpoint:

$\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$

Substitute:

$x=\dfrac{-8+(-4)}{2}=\dfrac{-12}{2}=-6\\\\y=\dfrac{-6+(-14)}{2}=\dfrac{-20}{2}=-10$

We have h = -6 and k = -10.

The radius is the distance between a center and the point on a circumference of a circle.

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute (-6, -10) and (-8, -6):

$r=\sqrt{(-8-(-6))^2+(-6-(-10))^2}=\sqrt{(-2)^2+4^2}=\sqrt{4+16}=\sqrt{20}$

Finally we have

$(x-(-6))^2+(y-(-10))^2=(\sqrt{20})^2\\\\(x+6)^2+(y+10)^2=20$

5 0

3 years ago

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What angles are supplementary?

vova2212 [387]

8 and 5 and I hate IXL ugh

5 0

3 years ago