The equation of the circle is x² + y² = 25 and the transformation equation of the circle is x² + y² = 100.
<h3>What is the equation of the circle?</h3>
The following parameters are derived from the question:
(x - a)² + (y - b)² = r²
Center (a, b) = (0, 0)
Radius (r) = 5
Then the equation of the circle will be
x² + y² = 5²
x² + y² = 25
Then the transformation of the circle will be
Center (a, b) = (0, 0)
Radius (r) = 10
Then the equation of the circle will be
x² + y² = 10²
x² + y² = 100
More about the equation of circle link is given below.
brainly.com/question/10618691
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P^2 -5p + 6 = 0
(p - 2)(p - 3) = 0
p - 2 = 0 p - 3 = 0
p = 2, p= 3
Answer:
12/3 = 4
Step-by-step explanation:
The answer is going to be the second one
You can use the definition:
![\displaystyle f'(x) = \lim_{h\to0}\frac{f(x+h)-f(x)}h](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%27%28x%29%20%3D%20%5Clim_%7Bh%5Cto0%7D%5Cfrac%7Bf%28x%2Bh%29-f%28x%29%7Dh)
Then if
![f(x) = 2x^2+3x-2](https://tex.z-dn.net/?f=f%28x%29%20%3D%202x%5E2%2B3x-2)
we have
![f(x+h) = 2(x+h)^2+3(x+h) - 2 = 2x^2 + 4xh+2h^2+3x+3h-2](https://tex.z-dn.net/?f=f%28x%2Bh%29%20%3D%202%28x%2Bh%29%5E2%2B3%28x%2Bh%29%20-%202%20%3D%202x%5E2%20%2B%204xh%2B2h%5E2%2B3x%2B3h-2)
Then the derivative is
![\displaystyle f'(x) = \lim_{h\to0}\frac{(2x^2+4xh+2h^2+3x+3h-2)-(2x^2+3x-2)}h \\\\ f'(x) = \lim_{h\to0}\frac{4xh+2h^2+3h}h \\\\ f'(x) = \lim_{h\to0}(4x+2h+3) = \boxed{4x+3}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%27%28x%29%20%3D%20%5Clim_%7Bh%5Cto0%7D%5Cfrac%7B%282x%5E2%2B4xh%2B2h%5E2%2B3x%2B3h-2%29-%282x%5E2%2B3x-2%29%7Dh%20%5C%5C%5C%5C%20f%27%28x%29%20%3D%20%5Clim_%7Bh%5Cto0%7D%5Cfrac%7B4xh%2B2h%5E2%2B3h%7Dh%20%5C%5C%5C%5C%20f%27%28x%29%20%3D%20%5Clim_%7Bh%5Cto0%7D%284x%2B2h%2B3%29%20%3D%20%5Cboxed%7B4x%2B3%7D)
I'm guessing the second part of the question asks you to find the tangent line to <em>f(x)</em> at the point <em>a</em> = 0. The slope of the tangent line to this point is
![f'(0) = 4(0) + 3 = 3](https://tex.z-dn.net/?f=f%27%280%29%20%3D%204%280%29%20%2B%203%20%3D%203)
and when <em>a</em> = 0, we have <em>f(a)</em> = <em>f</em> (0) = -2, so the graph of <em>f(x)</em> passes through the point (0, -2).
Use the point-slope formula to get the equation of the tangent line:
<em>y</em> - (-2) = 3 (<em>x</em> - 0)
<em>y</em> + 2 = 3<em>x</em>
<em>y</em> = 3<em>x</em> - 2