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2 years ago

A circle has a center at A (-3,4) and radius of 4 units. What is the equation of the circle? Show work!

Mathematics

1 answer:

Studentka2010 [4]2 years ago

7 0

Answer:

The equation of the circle is (x+3)^2 + (y-4)^2 = 16.

Step-by-step explanation:

The equation of a circle can be calculated if the center and the radius are known. The equation of a circle is given by (x-h)² + (y-k)² = r²

Where (h, k) = center coordinates and r = radius.

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Due tomorrow Please help!!

anastassius [24]

Answer:

$\frac{1}{8}$ = $\frac{b}{4^{2} }$

b = $\frac{1}{8}$ × $4^{2}$

b = $\frac{4^{2} }{8}$

then b = $\frac{c^{2} }{a}$ that is ORANGE

-2 - v = -7

v = -7 +2

then v = k + m that is BROWN

$\frac{14}{q}$ = $\frac{-7}{-4}$

q = $\frac{14 * -4 }{-7}$

then q = $\frac{ps}{r}$ that is YELLOW

4m + 2(n) = 5 (n)

4m = 5 (n) - 2n

4m = 3n

m = $\frac{3n}{4}$

then m = $\frac{3n}{4}$ that is RED

-8 = $\frac{x}{-5}$ - 6

$\frac{x}{-5}$ = -8 + 6

x = -5 (-8 + 6)

then x = y( w + z ) that is LIGHT GREEN

7 0

2 years ago

Find the coordinates of the centroid of triangle RST. SHOW YOUR WORK so I can see if the answer makes sense (no links)

zimovet [89]

Note the coordinates of each point: R(-4, 5), S(5, 1), T(2, -3).

The centroid is the point whose coordinates are the average of the coordinates of R, S, and T.

<em>x</em>-coordinate: (-4 + 5 + 2)/3 = 3/3 = 1

<em>y</em>-coordinate: (5 + 1 - 3)/3 = 3/3 = 1

So the centroid is (1, 1).

3 0

1 year ago

(ASAP PICTURE ADDED) What is the simplified form of the following expression?

bonufazy [111]

Answer:

option c is correct.

Step-by-step explanation:

$7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{16x}\right)-3\left(\sqrt[3]{8x}\right)$

WE need to simplify this equation.

Solve the parenthesis of each term.

$=7\left\sqrt[3]{2x}\right-3\left\sqrt[3]{16x}\right-3\left\sqrt[3]{8x}\right$

Now, We will find factors of the terms inside the square root

factors of 2: 2

factors of 16 : 2x2x2x2

factors of 8: 2x2x2

Putting these values in our equation: $=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2X2X2X2 x}\right)-3\left(\sqrt[3]{2X2X2 x}\right)\\=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2X2X2} \sqrt[3] {2 x}\right)-3\left(\sqrt[3]{2X2X2} \sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2^3} \sqrt[3] {2 x}\right)-3\left(\sqrt[3]{2^3} \sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2x}\right)-3*2\left(\sqrt[3] {2 x}\right)-3*2\left(\sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2}\sqrt[3]{x}\right)-6\left(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)$

Adding like terms we get:

$=7\left(\sqrt[3]{2}\sqrt[3]{x}\right)-6\left(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right\\=(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)\\$