Answer:
r = √13
Step-by-step explanation:
Starting with x^2+y^2+6x-2y+3, group like terms, first x terms and then y terms: x^2 + 6x + y^2 -2y = 3. Please note that there has to be an " = " sign in this equation, and that I have taken the liberty of replacing " +3" with " = 3 ."
We need to "complete the square" of x^2 + 6x. I'll just jump in and do it: Take half of the coefficient of the x term and square it; add, and then subtract, this square from x^2 + 6x: x^2 + 6x + 3^2 - 3^2. Then do the same for y^2 - 2y: y^2 - 2y + 1^2 - 1^2.
Now re-write the perfect square x^2 + 6x + 9 by (x + 3)^2. Then we have x^2 + 6x + 9 - 9; also y^2 - 1y + 1 - 1. Making these replacements:
(x + 3)^2 - 9 + (y - 1)^2 -1 = 3. Move the constants -9 and -1 to the other side of the equation: (x + 3)^2 + (y - 1)^2 = 3 + 9 + 1 = 13
Then the original equation now looks like (x + 3)^2 + (y - 1)^2 = 13, and this 13 is the square of the radius, r: r^2 = 13, so that the radius is r = √13.
Answer: The graph of f(x) is horizontally stretched
A. It's a composite function, so basically, wherever you see a p, replace it with 5t, because we are given that information. So, your answer is:
![A[p(t)] = 5t \pi 2=10t \pi](https://tex.z-dn.net/?f=A%5Bp%28t%29%5D%20%3D%205t%20%5Cpi%202%3D10t%20%5Cpi%20)
B. Let's use the function we created, and just plug in 2 for t:
![A[p(2)] = 10(2) \pi](https://tex.z-dn.net/?f=A%5Bp%282%29%5D%20%20%3D%2010%282%29%20%5Cpi%20)
![A[p(2)] = 62.83](https://tex.z-dn.net/?f=A%5Bp%282%29%5D%20%3D%2062.83)
So, your answer is
(approximately)
62.83 units².
If there is 20 apples to 10 pears. you could say this as a ratio of 20:10 and then simplify this to 2:1.
<span>This means for every 2 apples there is 1 pear.</span>
Answer:
A = pi r^2
Step-by-step explanation:
The formula for area of a circle is given by
A = pi r^2
