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Mashcka [7]
3 years ago
5

A diameter of a circle has endpoints p(-10,-2) and Q(4,6)

Mathematics
1 answer:
hodyreva [135]3 years ago
4 0

Answer:

a. (-3,2)

b. sqrt65

c.

Step-by-step explanation:

a. To find the center of the circle, you can think of it just like finding the midpoint between the two endpoints. To find a midpoint between two endpoints, you take the average of the x values to get the x coordinate, and you take the average of the y values to get the y coordinate of the midpoint. Therefore, if (-10, -2) is (x1, y1) and (4, 6) is (x2, y2), the midpoint/center of the circle would be:

( (x1+x2)/2, (y1+y2)/2 ). When you plug in our x and y values, you get (-3, 2).

b. To find the radius of a circle, you need to know the center/midpoint of the circle which we solved for in part a. The formula for finding the radius of a circle with the center is (x-h)^2 + (y-k)^2 = r^2 for (h, k) as the center. The coordinates of the center that we found earlier for this circle are (-3, 2). With that, we just plug in our numbers into the formula, and we get:

(x+3)^2 + (y-2)^2 = r^2. Now, to get r, we can choose one of the original two endpoints given and plug in the x and y coordinates from that point into this equation. I like (4, 6), so I'm going to plug in 4 for x and 6 for y, and so we get (4+3)^2 + (6-2)^2 = r^2 which equals 49 + 16 = r^2 when simplified. 49 plus 16 is equal to 65, so we get 65 = r^2. To finally get r, we square root both sides of the equation to get r =  sqrt65 which is already in the simplest radical form.

c. The circle equation is (x-h)^2 + (y-k)^2 = r^2, like I said in part b. Therefore, we already have our circle equation! We just plug in our center points and we get (x+3)^2 + (y-2)^2 = sqrt65. This is usually an equation a question will give you for a circle, and with this information, they will expect you to find the center (h, k) or the circle and it's radius, r.

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Se tiene un rectángulo cuya base mide el doble que su altura y su área es 12
irina [24]

Answer:

Part 1) The perimeter is  P=6\sqrt{6}\ cm

Part 2) The diagonal is d=\sqrt{30}\ cm

Step-by-step explanation:

<u><em>The question in English is</em></u>

You have a rectangle whose base is twice the height and its area is 12

square centimeters. Calculate the perimeter of the rectangle and its diagonal

step 1

Find the dimensions of rectangle

we know that

The area of rectangle is equal to

A=bh

A=12\ cm^2

so

bh=12 ----> equation A

The base is twice the height

so

b=2h ----> equation B

substitute equation B in equation A

(2h)h=12\\2h^2=12\\h^2=6\\h=\sqrt{6}\ cm

Find the value of b

b=2\sqrt{6}\ cm

step 2

Find the perimeter of rectangle

The perimeter is given by

P=2(b+h)

substitute

P=2(2\sqrt{6}+\sqrt{6})\\P=6\sqrt{6}\ cm

step 3

Find the diagonal of rectangle

Applying the Pythagorean Theorem

d^2=b^2+h^2

substitute

d^2=(2\sqrt{6})^2+(\sqrt{6})^2

d^2=30

d=\sqrt{30}\ cm

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what TRANSFORMATIONS take place to the parent graph f(x)= log(x) to achieve the graph g(x)=log(-2x-4)+5?
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