Answer:
( x + 2 )² + ( y - 4 )² = 36
Step-by-step explanation:
Standard equation of a circle - ( x - h )² + ( y - k )² = r²
where r = radius and center is at ( h , k )
The given center is ( -2 , 4 )
Thus h = -2 and k = 4
and the given radius is 6
Thus, r = 6
that being said, we plug in the given values into the standard equation
( x - ( - 2 )² + ( y - 4 )² = 6²
Finally we simplify
In - ( -2 ) the two negative signs cancel out and it turns into + 2
6² = 36
The final equation of the circle is ( x + 2 )² + ( y - 4 )² = 36
Perpendicular Transversal Theorem <span>can be used to prove that d is perpendicular to t.
The theorem states that : </span><span>In a plane, if a line is perpendicular to one of the 2 parallel lines then it is perpendicular to the other.</span>
![\frac{3x-5}{x-5}](https://tex.z-dn.net/?f=%20%5Cfrac%7B3x-5%7D%7Bx-5%7D%20)
> 0
First, note that x is undefined at 5. / x ≠ 5
Second, replace the inequality sign with an equal sign so that we can solve it like a normal equation. / Your problem should look like:
![\frac{3x-5}{x-5}](https://tex.z-dn.net/?f=%20%5Cfrac%7B3x-5%7D%7Bx-5%7D%20)
= 0
Third, multiply both sides by x - 5. / Your problem should look like: 3x - 5 = 0
Forth, add 5 to both sides. / Your problem should look like: 3x = 5
Fifth, divide both sides by 3. / Your problem should look like: x =
Sixth, from the values of x above, we have these 3 intervals to test:
x <
![\frac{5}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B5%7D%7B3%7D%20)
![\frac{5}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B5%7D%7B3%7D%20)
< x < 5
x > 5
Seventh, pick a test point for each interval.
1. For the interval x <
![\frac{5}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B5%7D%7B3%7D%20)
:
Let's pick x - 0. Then,
![\frac{3x0-5}{0-5}](https://tex.z-dn.net/?f=%20%5Cfrac%7B3x0-5%7D%7B0-5%7D%20)
> 0
After simplifying, we get 1 > 0 which is true.
Keep this interval.
2. For the interval
![\frac{5}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B5%7D%7B3%7D%20)
< x < 5:
Let's pick x = 2. Then,
![\frac{3x2-5}{2-5}](https://tex.z-dn.net/?f=%20%5Cfrac%7B3x2-5%7D%7B2-5%7D%20)
> 0
After simplifying, we get -0.3333 > 0, which is false.
Drop this interval.
3. For the interval x > 5:
Let's pick x = 6. Then,
![\frac{3x6-5}{6-5}](https://tex.z-dn.net/?f=%20%5Cfrac%7B3x6-5%7D%7B6-5%7D%20)
> 0
After simplifying, we get 13 > 0, which is ture.
Keep this interval.
Eighth, therefore, x <
![\frac{5}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B5%7D%7B3%7D%20)
and x > 5
Answer: x <
![\frac{5}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B5%7D%7B3%7D%20)
and x > 5