If the center is in point P(a,b) and has a radius r then circle equation is:
(x-a)²+(y-b)²=r².
In your task we've got P(0,0) so the circle equation is x²+y²=r².
We have to find r. Radius start at (0,0) and end at (-4,-6) therefore
r=√((-4)²+(-6)²) and
r²=(-4)²+(-6)²=16+36=52.
So answer is x²+y²=52.
Answer:
Step-by-step explanation:
(a + b)² =a² + 2ab + b²
(a -b)² = a² - 2ab + b²
1) y = (x -1)²
y= x² - 2*x*1 + 1
y = x² - 2x + 1
Ans: C
2)y = (x +4)² + 5
y = x² +2*x*4 + 4² + 5
= x² + 8x + 16 + 5
y = x² + 8x + 21
C
3) y = -(x + 9)²- 10
y = - [x² + 18x + 81] - 10
= -x² - 18x - 81 - 10
y =-x² - 18x - 91
B
4) y = 3(x + 2)² - 18
y =3 [x² + 4x + 4] - 18
y = 3x² + 12x + 12 - 18
y =3x² + 12x - 6
A
5) y = -2(x + 1)² - 16
= -2[x² + 2x + 1] -16
= -2x² - 4x - 2 - 16
y = -2x² - 4x - 18
A
6) y = 5(x + 5)²
=5[x²+ 10x + 25]
y = 5x² +50x + 125
A
7)y = (1/2)(x + 8)² - 8
y = (1/2) (x² + 16x + 64) - 8

A
8) y = (x + 3/2)² + 3/4

C
9) y = 2[x² + 16x + 64] - 5x
y = 2x² + 32x + 64 - 5x
y =2x² + 27x + 6
Answer:
Step-by-step explanation:
if im not wrong it should be the first one
Answer:
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Step-by-step explanation:
Since we have a cubic root, we're interested in factoring cubes inside the root, so that we can take them out. If we factor 648, we have

So, we have
![3x\sqrt[3]{648 x^4 y^8} = \sqrt[3]{3\times 6^3\cdot x^3\cdot x \cdot y^6\cdot y^2}=3x\cdot 6\cdot x\cdot y^2\sqrt[3]{3\cdot x\cdot y^2}](https://tex.z-dn.net/?f=3x%5Csqrt%5B3%5D%7B648%20x%5E4%20y%5E8%7D%20%3D%20%5Csqrt%5B3%5D%7B3%5Ctimes%206%5E3%5Ccdot%20x%5E3%5Ccdot%20x%20%5Ccdot%20y%5E6%5Ccdot%20y%5E2%7D%3D3x%5Ccdot%206%5Ccdot%20x%5Ccdot%20y%5E2%5Csqrt%5B3%5D%7B3%5Ccdot%20x%5Ccdot%20y%5E2%7D)
And the result simplifies to
![18x^2y^2\sqrt[3]{3xy^2}](https://tex.z-dn.net/?f=18x%5E2y%5E2%5Csqrt%5B3%5D%7B3xy%5E2%7D)