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3 years ago

Find the other endpoint of the line segment with the given endpoint and midpoint.

Mathematics

1 answer:

VLD [36.1K]3 years ago

8 0

Find the distance between the two x values: -10 to -7 is 3 units, so add to the midpoint: -7 + 3 = -4

The x value is -4

Now do the same for y:

7 to -3 is 10 units

Because the midpoint is less than the end point subtract:

Y = -3 -10 = -13

Y = -13

The endpoint would be (-4, -13)

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What is the slope of the line that passes through the points (-10, -8) and

RoseWind [281]

Answer:

Equation: y = -4x + 32

Slope: -4

Step-by-step explanation:

Given : (-10,-8) and (-8, -16)

Slope Formula: $\frac{y_1 - y_2}{x_1-x_2}$

Solve For Slope, Input Given Points:

$\frac{-8 - (-16)}{-10 - (-8)}$
$\frac{8}{-2}$
$-4$

Solve for y-intercept:

Input the slope into the equation y = mx + b. Plugin x and y values for the x and y variables to find b.

y = -4x + b
-8 = -10(4) + b
-8 = -40 + b
b = 32

Equation of LIne: y = -4x + 32

-Chetan K

5 0

2 years ago

Read 2 more answers

Please help me with #'s 5&6

Fynjy0 [20]

$5.\\(x+3)^2+7=-2\qquad|-7\\\\(x+3)^2=-9 < 0\\\\\text{NO REAL SOLUTION}\\\\(x+3)^2=-9\to x+3=\pm\sqrt{-9}\\\\x+3=-3i\ \vee\ x+3=3i\qquad|-3\\\\\boxed{x=-3-3i\ \vee\ x=-3+3i}$

$6.\\4(x-10)^2=25\qquad|:4\\\\(x-10)^2=\dfrac{25}{4}\to x-10=\pm\sqrt{\dfrac{25}{4}}\\\\x-10=-\dfrac{\sqrt{25}}{\sqrt4}\ \vee\ x-10=\dfrac{\sqrt{25}}{\sqrt4}\\\\x-10=-\dfrac{5}{2}\ \vee\ x-10=\dfrac{5}{2}\\\\x-10=-2.5\ \vee\ x-10=2.5\qquad|+10\\\\\boxed{x=7.5\ \vee\ x=12.5}$

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3 years ago

What is the reciprocal of 159?

wariber [46]

<h2>Hey there! </h2>

<h2>The Reciprocal of 159 is:</h2>

$\frac{1}{159}$

${where \: 1 \: is \: hidden}$

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3 years ago

What's (7x/2-5x+3)+(2x/2+4x-6)

Anika [276]

I'm assuming that when you wrote "(7x/2-5x+3)+(2x/2+4x-6)," you actually meant "<span>(7x^2-5x+3)+(2x^2+4x-6). Correct me if I'm wrong here.

</span><span>+(7x^2-5x+3)
</span><span>+(2x/2+4x-6)
-------------------
=9x^2 - x - 3 (answer) </span>

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