Write an equation to f the circle with the center (-5,7) and a radius of 3

Equation of a circle: (x - h)² + (y - k)² = r²

(h, k) = center

r = radius

We have the following values:

(-5, 7) = center (h, k)

3 = radius (r)

~Substitute with the given values

(x - (-5))² + (y - (7))² = (3)²

~Simplify

(x + 5)² + (y - 7)² = 9

(x + 5)² + (y - 7)² = 9

Best of Luck!

Explanation: The formula for the equation of a circle is shown below.

(x - h)² + (y - k)² = r²

In the equation above, (h, k) is the center the circle and r is the radius.

Since we are given that the center is (-5, 7) which represents (h, k) and the radius is 3 which represents r, we simply plug these values into the formula and we have [x - (-5)]² + [y - (7)]² = (3)².

Notice that I changed the parentheses that were in the original formula to brackets so that we wouldn't have too many sets of parentheses when plugging our values into the formula.

Finally, we simplify inside the brackets.

[x - (-5)] simplifies to [x + 5].

So we have (x + 5)² + (y - 7)² = 9.

Notice that I changed the brackets back to

parentheses in the final answer.

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.

i dont know men

step-by-step explanation: