Answer: f(120°) = (√3) + 1/2
Step-by-step explanation:
i will solve it with notable relations, because using a calculator is cutting steps.
f(120°) = 2*sin(120°) + cos(120°)
=2*sin(90° + 30°) + cos(90° + 30°)
here we can use the relations
cos(a + b) = cos(a)*cos(b) - sin(a)*sin(b)
sin(a + b) = cos(a)*sin(b) + cos(b)*sin(a)
then we have
f(120°) = 2*( cos(90°)*sin(30°) + cos(30°)*sin(90°)) + cos(90°)*cos(30°) - sin(90°)*sin(30°)
and
cos(90°) = 0
sin(90°) = 1
cos(30°) = (√3)/2
sin(30°) = 1/2
We replace those values in the equation and get:
f(120°) = 2*( 0 + (√3)/2) + 0 + 1/2 = (√3) + 1/2
Yes it will, they are the same value just in diff locations
Hi the answer to your question is
((3y - 3.5) / (3y + 6)) * 25.5 = (17
/ 25.5) * 25.5
((3y - 3.5) / (3y + 6)) * 25.5 = 17
(76.5y - 89.25) / (3y + 6) = 17
((76.5y - 89.25) / (3y + 6)) * (3y +
6) = 17 * (3y + 6) 76.5y - 89.25 = 51y + 102 76.5y - 51y = 102 + 89.25
25.5y = 191.25
y = 191.25 / 25.5
<span>y = 7.5</span>
Answer:
The probability of selecting a black card or a 6 = 7/13
Step-by-step explanation:
In this question we have given two events. When two events can not occur at the same time,it is known as mutually exclusive event.
According to the question we need to find out the probability of black card or 6. So we can write it as:
P(black card or 6):
The probability of selecting a black card = 26/52
The probability of selecting a 6 = 4/52
And the probability of selecting both = 2/52.
So we will apply the formula of compound probability:
P(black card or 6)=P(black card)+P(6)-P(black card and 6)
Now substitute the values:
P(black card or 6)= 26/52+4/52-2/52
P(black card or 6)=26+4-2/52
P(black card or 6)=30-2/52
P(black card or 6)=28/52
P(black card or 6)=7/13.
Hence the probability of selecting a black card or a 6 = 7/13 ....
No. It doesn’t, the square root is different than the approximate length.
