Use the trig identity
2*sin(A)*cos(A) = sin(2*A)
to get
sin(A)*cos(A) = (1/2)*sin(2*A)
So to find the max of sin(A)*cos(A), we can find the max of (1/2)*sin(2*A)
It turns out that sin(x) maxes out at 1 where x can be any expression you want. In this case, x = 2*A.
So (1/2)*sin(2*A) maxes out at (1/2)*1 = 1/2 = 0.5
The greatest value of sin(A)*cos(A) is 1/2 = 0.5
Answer:
6(6a-2b)
Step-by-step explanation:
Answer:
C..... I think it is a answer
For this case we have to, by defining properties of powers and roots the following is fulfilled:
![\sqrt [n] {a ^ m} = a ^ {\frac {m} {n}}](https://tex.z-dn.net/?f=%5Csqrt%20%5Bn%5D%20%7Ba%20%5E%20m%7D%20%3D%20a%20%5E%20%7B%5Cfrac%20%7Bm%7D%20%7Bn%7D%7D)
We must rewrite the following expression:
![\sqrt [3] {8 ^ {\frac {1} {4} x}}](https://tex.z-dn.net/?f=%5Csqrt%20%5B3%5D%20%7B8%20%5E%20%7B%5Cfrac%20%7B1%7D%20%7B4%7D%20x%7D%7D)
Applying the property listed we have:
![\sqrt [3] {8 ^ {\frac {1} {4} x}} = 8 ^ {\frac{\frac {1} {4} x} {3} }= 8 ^ {\frac {1} {4 * 3} x} = 8 ^ {\frac {1} {12} x}](https://tex.z-dn.net/?f=%5Csqrt%20%5B3%5D%20%7B8%20%5E%20%7B%5Cfrac%20%7B1%7D%20%7B4%7D%20x%7D%7D%20%3D%208%20%5E%20%7B%5Cfrac%7B%5Cfrac%20%7B1%7D%20%7B4%7D%20x%7D%20%7B3%7D%20%7D%3D%208%20%5E%20%7B%5Cfrac%20%7B1%7D%20%7B4%20%2A%203%7D%20x%7D%20%3D%208%20%5E%20%7B%5Cfrac%20%7B1%7D%20%7B12%7D%20x%7D)
Using the property again we have to:
![8 ^ {\frac {1} {12} x} = \sqrt [12] {8 ^ x}](https://tex.z-dn.net/?f=8%20%5E%20%7B%5Cfrac%20%7B1%7D%20%7B12%7D%20x%7D%20%3D%20%5Csqrt%20%5B12%5D%20%7B8%20%5E%20x%7D)
Thus, the correct option is option C
Answer:
Option C
To find the mean, you need to find the average. The equation that I used was 61(1)+64(3)+67(5)+70(6)+73(5) and then divided that total by the total number of 20.
The answer would be D) 68.5
