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Anna [14]
3 years ago
6

Please answer this question fast in 2 minutes

Mathematics
1 answer:
Stolb23 [73]3 years ago
5 0

Answer:

(-20, 19)

Step-by-step explanation:

We need to use the Midpoint Formula, which says that given two points (x_1,y_1) and (x_2,y_2), the midpoint is (\frac{x_1+x_2}{2} ,\frac{y_1+y_2}{2} ).

Here, we are given one of the endpoints and the midpoint:

endpoint S = (5, -8)

midpoint M = (-7.5, 5.5)

Plug these into the formula. In \frac{x_1+x_2}{2}, x1 = 5, and this whole expression is equal to -7.5:

\frac{x_1+x_2}{2}=\frac{5+x_2}{2} =-7.5

Solve for x2:

5 + x2 = 2 * (-7.5) = -15

x2 = -15 - 5 = -20

Now, let's find y2. y1 is just -8, and the entire expression for the y-coordinate of the midpoint is equal to 5.5. So:

\frac{y_1+y_2}{2}=\frac{-8+y_2}{2} =5.5

Solve for y2:

-8 + y2 = 2 * 5.5

-8 + y2 = 11

y2 = 11 + 8 = 19

The coordinates of T are thus (-20, 19).

<em>~ an aesthetics lover</em>

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Determine whether the three points are collinear. (3, -10), (-2, -7), (0, -5)
Mumz [18]

The points are not collinear.

Solution:

Let A, B and C be (3, -10), (-2, -7) and (0, -5).

<em>If slopes of any two points are same, then the points are collinear.</em>

Slope formula:

$m=\frac{y_2-y1}{x_2-x_1}

<u>Slope of AB:</u>

$m_1=\frac{-7-(-10)}{-2-3}

$m_1=\frac{-7+10}{-5}

$m_1=-\frac{3}{5}

<u>Slope of BC:</u>

$m_2=\frac{-5-(-7)}{0-(-2)}

$m_2=\frac{-5+7}{2}

$m_2=\frac{2}{2}

m_2=1

<u>Slope of CA:</u>

$m_3=\frac{-10-(-5)}{3-0}

$m_3=\frac{-10+5}{3}

$m_3=-\frac{5}{3}

m_1\neq m_2 \neq m_3

Slope of AB ≠ Slope of BC ≠ Slope of AC

Therefore the points are not collinear.

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Step-by-step explanation:

One digit is given as 8, and the other digit is half of 8, or 4.

The number 48 has 8 as a factor; 84 does not.

The product is 48.

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