Answer:
(2x-3) (2x+3)
zeros, x intercepts: -3/2, 3/2
Step-by-step explanation:
4x^2 -9
We know the difference of squares is a^2 -b^2
This factors into (a-b) (a+b)
Let 4x^2 =a^2
Taking the square root
2x =a
Let b^2 =9
Taking the square root
b= 3
(4x^2-9 ) = (2x-3) (2x+3)
To find the zeros, we set the equation equal to zero
(4x^2-9 ) = (2x-3) (2x+3) =0
Using the zero product property
2x-3 =0 and 2x+3 =0
2x-3+3 = 0+3 2x+3-3 = 0-3
2x=3 2x=-3
Divide by 2
2x/2 = 3/2 2x/2 = -3/2
x = 3/2 x = -3/2
These are the zeros of the equation (which are also the x intercepts)
Approximately 0.3246 (rounded to the nearest ten thousandth). Find the square root of 40, then subtract 6.
8.77 x 10^1 i believe but would double check
4/5 ,8/10, 16/20,
You multiply both numbers with the same number.
For ex. 4×2=8 5×2=10 so we have 8/10
Since 3x+17=5x-3, x= 10. so angles
