Your answer is:
B. subtract 10 to both sides
Hope it helps
Have a great day :)
Let me know if it’s correct
The cosine of an angle is the x-coordinate of the point where its terminal ray intersects the unit circle. So, we can draw a line at x=-1/2 and see where it intersects the unit circle. That will tell us possible values of θ/2.
We find that vertical line intersects the unit circle at points where the rays make an angle of ±120° with the positive x-axis. If you consider only positive angles, these angles are 120° = 2π/3 radians, or 240° = 4π/3 radians. Since these are values of θ/2, the corresponding values of θ are double these values.
a) The cosine values repeat every 2π, so the general form of the smallest angle will be
... θ = 2(2π/3 + 2kπ) = 4π/3 + 4kπ
b) Similarly, the values repeat for the larger angle every 2π, so the general form of that is
... θ = 2(4π/3 + 2kπ) = 8π/3 + 4kπ
c) Using these expressions with k=0, 1, 2, we get
... θ = {4π/3, 8π/3, 16π/3, 20π/3, 28π/3, 32π/3}
![\bf \textit{difference and sum of cubes} \\\\ a^3+b^3 = (a+b)(a^2-ab+b^2) ~\hfill a^3-b^3 = (a-b)(a^2+ab+b^2) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \boxed{a^6+b^6}\implies a^{2\cdot 3}+b^{2\cdot 3}\implies (a^2)^3+(b^2)^3 \\[2em] [a^2+b^2] [(a^2)^2-a^2b^2+(b^2)^2]\implies \boxed{(a^2+b^2)(a^4-a^2b^2+b^4)}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bdifference%20and%20sum%20of%20cubes%7D%20%5C%5C%5C%5C%20a%5E3%2Bb%5E3%20%3D%20%28a%2Bb%29%28a%5E2-ab%2Bb%5E2%29%20~%5Chfill%20a%5E3-b%5E3%20%3D%20%28a-b%29%28a%5E2%2Bab%2Bb%5E2%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cboxed%7Ba%5E6%2Bb%5E6%7D%5Cimplies%20a%5E%7B2%5Ccdot%203%7D%2Bb%5E%7B2%5Ccdot%203%7D%5Cimplies%20%28a%5E2%29%5E3%2B%28b%5E2%29%5E3%20%5C%5C%5B2em%5D%20%5Ba%5E2%2Bb%5E2%5D%20%5B%28a%5E2%29%5E2-a%5E2b%5E2%2B%28b%5E2%29%5E2%5D%5Cimplies%20%5Cboxed%7B%28a%5E2%2Bb%5E2%29%28a%5E4-a%5E2b%5E2%2Bb%5E4%29%7D)
about the second one... well, is a "fait accompli" that using the pythagorean theorem, if x = 8 and y = 5, the hypotenuse must be √(8² + 5²) = √(89), which is neither of those choices.
5, 8, 13 are no dice, namely 5² + 8² ≠ 13
25, 64, 17 is are no dice too, because 25² + 17² ≠ 64²
however, 5,12 and 13 are indeed a pythagorean triple
also is 39, 80, 89.
when looking for a pythagorean triple, recall that c² = a² + b².
so the longest leg is the sum of the square of the small ones.
so what you'd do is, check the small legs, square them, add them up, if they're indeed a pythagorean triple, they "must" add up to the longest leg.
Answer:

when x is 2, y is -4

None of the objectives is right
OK so my answer is from what you’ve just explained that means you multiply the six and the 114 is 684 so that means if you have to do that to 13 of them that means 684×13 is what your equation would be in the answer to 684×13 is 8892