Answer:
c
Step-by-step explanation:
Answer:
W'(4, -2), X'(3, -4), Y'(1, -4), Z'(0, -2)
I hope this helps you. If you can give brainliest that would be greatly appreciated.
Solve for x over the real numbers:
3 x^2 + 5 x + 1 = 0
Hint: | Using the quadratic formula, solve for x.
x = (-5 ± sqrt(5^2 - 4×3))/(2×3) = (-5 ± sqrt(25 - 12))/6 = (-5 ± sqrt(13))/6:
Answer: x = (-5 + sqrt(13))/6 or x = (-5 - sqrt(13))/6
\left[x \right] = \left[ 3\right][x]=[3] totally answer
Answer:

Step-by-step explanation:
You can do this using the quadratic equation:

Let's first setup our expression:
4x² + 12x = 135
the quadratic formula is:
ax² + bx + c = 0
4x² + 12x - 135 = 0
Then:
a = 4
b = 12
c = -135
Now we plug in our coefficients and solve:

We solve for both to determine the positive one:





So if you are looking for a positive solution, just take the positive one as x.