Suppose y=0. That means
<span>x+z = 18 </span>
<span>x-z = 26/5 </span>
<span>x = 58/5 </span>
<span>z = 32/5 </span>
<span>So, the largest x can be is 58/5 </span>
<span>Now, suppose we toss in y. If we subtract y/2 from both x and z, then the difference between x and z is still 26/5, and the sum is </span>
<span>(x-y/2) + y + (z-y/2) = x+z = 18 </span>
<span>For example, if y = 4, </span>
<span>x = 48/5 </span>
<span>y = 20/5 </span>
<span>z = 22/5 </span>
<span>x-z is still 26/5, and the sum is still 18. </span>
Answer:
D is the answer
Step-by-step explanation:
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
The correct answer is C
Good luck on the rest :D