You may notice that all of the terms have a common factor:
. So, let's take the
out of all the terms. This gives us:
![x(x^2 + 2x - 24) = 0](https://tex.z-dn.net/?f=x%28x%5E2%20%2B%202x%20-%2024%29%20%3D%200)
Using the Zero Product Property, we can say that both
and
.
Let's solve both.
is obvious, but we still need to solve our other term:
![x^2 + 2x - 24 = 0](https://tex.z-dn.net/?f=x%5E2%20%2B%202x%20-%2024%20%3D%200)
![(x + 6)(x - 4) = 0](https://tex.z-dn.net/?f=%28x%20%2B%206%29%28x%20-%204%29%20%3D%200)
![x = -6, 4](https://tex.z-dn.net/?f=x%20%3D%20-6%2C%204)
- Apply the Zero Product Property
Our answer is x = -6, 0, 4.
Answer:
The perpendicular slope is -4/3
Step-by-step explanation:
Perpendicular slopes multiply to -1
m * 3/4 = -1
Multiply each side by 4/3
m * 3/4 *4/3 = -1 * 4/3
m = -4/3
The perpendicular slope is -4/3
They are not equivalent because 27-(63-8) is equal to -28 and (72-63)-8 is equal to 1.<span />
Answer:
T' ![(\frac{5}{13},- \frac{12}{13})](https://tex.z-dn.net/?f=%28%5Cfrac%7B5%7D%7B13%7D%2C-%20%5Cfrac%7B12%7D%7B13%7D%29)
Step-by-step explanation:
See the diagram attached.
This is a unit circle having a radius (r) = 1 unit.
So, the length of the circumference of the circle will be 2πr = 2π units.
Now, the point on the circle at a distance of x along the arc from P is T
.
Therefore, the point on the circle at a distance of 2π - x along the arc from P will be T'
, where, T' is the image of point T, when reflected over the x-axis. (Answer)
y = -13, 13
There are two answers because you are taking the square root of a number. If we go the other way, to check our work, (-13)^2 = 169 and (13)^2 = 169.
Hope this helps! :)