What is the radius of the circle given by the equation below?

Mathematics

1 answer:

grandymaker [24]3 years ago

8 0

$x^2+18x+y^2+18y=-146 \\ \\( x^2+18x + 81)+(y^2+18y+81)-81-81=-146\\ \\(x+9)^2+(y+9)^2=-146+81+81\\ \\(x+9)^2+(y+9)^2=-146+162\\ \\(x+9)^2+(y+9)^2=16$

$This \ is \ in \ the \ general \ circle \ form : (x-h)^ 2+ (y-k)^2 = r^2, so: \\ \\ r^2=16 \\ \\r=\sqrt{16}\\ \\r=4 \\ \\ Answer : \ C. 4$

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Answer:

Part A: D. $y = (x + 1)(x+3)\\$

Part B: C. $3x(x-3) = 0\\$

Step-by-step explanation:

For part A) we just have to plug in 0 for x and solve for y until we find the equation that says 3 is the value for y when x is 0. For purposes of speeding up the process the correct answer is D. I will show how to check for it now.

The equation: $y = (x + 1)(x+3)$

Now plug in 0 for x.

$y = (0 + 1)(0+3)\\$

Now solve.

y = (1)(3)

y = 3

This proves that this is the correct answer.

For part B) we just have to plug in the give values for x separately and check for each value of x that it equals 0. For the purpose of speeding up the process the correct answer is C. I will show how to check for it now.

The equation: $3x(x-3) = 0$