Part 1
Part 2
Part 3
Part 4
Part 5
![[(2-3)^2-5]\cdot4=[(-1)^2-5]\cdot4 \\ \\ =(1-5)\cdot4=-4\cdot4=-16](https://tex.z-dn.net/?f=%5B%282-3%29%5E2-5%5D%5Ccdot4%3D%5B%28-1%29%5E2-5%5D%5Ccdot4%20%5C%5C%20%20%5C%5C%20%3D%281-5%29%5Ccdot4%3D-4%5Ccdot4%3D-16)
Part 6
Part 7
Part 8
Part 9
![(8-2)^2+\frac{1}{4}[4-3(6-10)]=6^2+\frac{1}{4}[4-3(-4)] \\ \\ =36+\frac{1}{4}[4-(-12)]=36+\frac{1}{4}(4+12)=36+\frac{1}{4}(16) \\ \\ =36+4=40](https://tex.z-dn.net/?f=%288-2%29%5E2%2B%5Cfrac%7B1%7D%7B4%7D%5B4-3%286-10%29%5D%3D6%5E2%2B%5Cfrac%7B1%7D%7B4%7D%5B4-3%28-4%29%5D%20%5C%5C%20%20%5C%5C%20%3D36%2B%5Cfrac%7B1%7D%7B4%7D%5B4-%28-12%29%5D%3D36%2B%5Cfrac%7B1%7D%7B4%7D%284%2B12%29%3D36%2B%5Cfrac%7B1%7D%7B4%7D%2816%29%20%5C%5C%20%20%5C%5C%20%3D36%2B4%3D40)
Part 10
Part 11
![3-2[2^3+(-1)^3]=3-2[8+(-1)] \\ \\ =3-2(8-1)=3-2(7)=3-14 \\ \\ -11](https://tex.z-dn.net/?f=3-2%5B2%5E3%2B%28-1%29%5E3%5D%3D3-2%5B8%2B%28-1%29%5D%20%5C%5C%20%20%5C%5C%20%3D3-2%288-1%29%3D3-2%287%29%3D3-14%20%5C%5C%20%20%5C%5C%20-11)
Part 12
Part 2
Part 3
Part 4
Part 5
Part 6
Part 7
Part 8
Part 9
Part 10
Part 11
Part 12
What is the result when (2x-5)^2 is subtracted from 8x-6?
1 answer:
You might be interested in
Let z = sin(x). This means z^2 = (sin(x))^2 = sin^2(x). This allows us to go from the equation you're given to this equation: 7z^2 - 14z + 2 = -5
That turns into 7z^2 - 14z + 7 = 0 after adding 5 to both sides. Use the quadratic formula to solve for z. The only solution is z = 1 (see attached image). Since we made z = sin(x), this means sin(x) = 1. All solutions to this equation will be in the form x = (pi/2) + 2pi*n, which is the radian form of the solution set. If you need the degree form, then it would be x = 90 + 360*n
The 2pi*n (or 360*n) part ensures we get every angle coterminal to pi/2 radians (90 degrees), which captures the entire solution set.
Note: The variable n can be any integer.
