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4 years ago

Write an equation for a circle with a diameter that has endpointsat (-4, -7) and (-2 ,-5)

Mathematics

1 answer:

sergiy2304 [10]4 years ago

7 0

Answer:

(x + 3)^2 + (y + 6)^2 = (√2)^2 = 2

Step-by-step explanation:

The standard equation for a circle with center at (h, k) and radius r is

(x - h)^2 + (y - k)^2 = r^2. We need to find h, k and r for this particular circle.

The length of the radius is half the length of the diameter. Find the length of this diameter first, and than halve the result.

diameter^2 = 2^2 + 2^2, or diameter^2 = 8, or diameter = √8 = 2√2.

This is because x changes from -4 to -2 (+2) and y changes from -7 to -5 (+2).

Then the radius is r = 2√2.

The coordinates of the center are found using the midpoint formula:

-4-2

xcenter = -------- = -3

-7-5

ycenter = -------- = - 6

Inserting -3 for h, -6 for k and √2 for r, we get:

(x + 3)^2 + (y + 6)^2 = (√2)^2 = 2

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Find the slope of the line through these pair of points.<br><br> (12,-18) (-15,-18)

zysi [14]

The slope, m, is given by the equation m = (y2-y1)/(x2-x1), where (x1, y1) and (x2, y2) are points on the graph. Thus:
m = (-18-(-18))/(-15-12)
= (-18+18)/-27
= 0/27 = 0
Therefor the slope is 0

8 0

3 years ago

What is:<br> <img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B4x%5E%7B2%7D%20-20x%2B25%20" id="TexFormula1" title=" \sqrt{4x^{2} -

Firdavs [7]

We are given :

$\sqrt{4x^{2}-20x+25}$

Step 1: factor the part inside square root

The function given inside square root is of quadratic form.

So let us try to factorise it using AC method.

Here A*C = 4*25 = 100

so we have to find factors of 100 that add up to give -20.

the two factors are -10 and -10.

Rewriting the function :

$4x^{2} -20x+25$

= $4x^{2} -10x-10x+25$

= $2x(2x-5) - 5(2x-5)$

= $(2x-5)^{2}$

Step 2:

Now we take square root of the factorised form

$\sqrt{(2x-5)^2}$

= $2x-5$

Answer : (2x-5)

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3 years ago

On a coordinate plane, triangle T V W has points (negative 4, 8), (0, 4), and (4, 4).

Vanyuwa [196]

The coordinates of the endpoints of line segments T'V' are; T'(-1, 2) and V'(0, 1).

<h3>What are the coordinates of the endpoints of the segment T'V'?</h3>

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On this note, given that the coordinates of T and V from the task content are; (-4, 8) and (0,4), it follows that the coordinates of the endpoints as required are; T'(-1, 2) and V'(0, 1).