Given the equation 4(3b + 2)² = 64,
dividing both sides of the equation by 4, we have
(3b + 2)² = 16 and getting the square root of both sides,
(3b + 2) = 4 and (3b + 2) = -4
We can solve for b for each equation and have
3b = 2 | 3b = -6
b = 2/3 | b = -2
Therefore, the values of b are 2/3 and -2 and from the choices, the answer is <span>A: b = 2/3 and b = -2.</span>
Answer:
because they are both in the circle
Step-by-step explanation:
Step-by-step explanation:

Step-by-step explanation:

According to this trigonometric function, −C gives you the OPPOSITE terms of what they really are, so be EXTREMELY CAREFUL:
![\displaystyle Phase\:[Horisontal]\:Shift → \frac{0}{\frac{1}{7}} = 0 \\ Period → \frac{2}{1}π = 2π](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Phase%5C%3A%5BHorisontal%5D%5C%3AShift%20%E2%86%92%20%5Cfrac%7B0%7D%7B%5Cfrac%7B1%7D%7B7%7D%7D%20%3D%200%20%5C%5C%20Period%20%E2%86%92%20%5Cfrac%7B2%7D%7B1%7D%CF%80%20%3D%202%CF%80)
Therefore we have our answer.
Extended Information on the trigonometric function
![\displaystyle Vertical\:Shift → D \\ Phase\:[Horisontal]\:Shift → \frac{C}{B} \\ Period → \frac{2}{B}π \\ Amplitude → |A|](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Vertical%5C%3AShift%20%E2%86%92%20D%20%5C%5C%20Phase%5C%3A%5BHorisontal%5D%5C%3AShift%20%E2%86%92%20%5Cfrac%7BC%7D%7BB%7D%20%5C%5C%20Period%20%E2%86%92%20%5Cfrac%7B2%7D%7BB%7D%CF%80%20%5C%5C%20Amplitude%20%E2%86%92%20%7CA%7C)
NOTE: Sometimes, your <em>vertical shift</em> might tell you to shift your graph below or above the <em>midline</em> where the amplitude is.
I am joyous to assist you anytime.