Answer:

Step-by-step explanation:

![\textsf{x=1/2[(4x+5)-50]}](https://tex.z-dn.net/?f=%5Ctextsf%7Bx%3D1%2F2%5B%284x%2B5%29-50%5D%7D)







ΔAOB is a right angled triangle. Therefore the Pythagorean Theorem applies in this situation.
θ is the angle from a standard position of the line OA
The length of the y component is √(1-0)2 +(-3-(-3))2] =√(12+ 02) = 1 A(-3,1) to B(-3,0) which is opposite
Then the length of the x-component is √[(-3-0)2 +(0-0)2] = √(9+0)= 3 B(-3,0) to O(0,0) which is adjacent
The length of vector OA is √[(-3-0)2 + (1-0)2] = √(9+1) = √(10) A(-3,1) to O(0,0) which is the hypotenuse of the triangle
θ = 180 - α
sinθ = sin(180-α) = opposite/hypotenuse = 1/√10
cosθ = adjacent/hypotenuse = -3/√10
tanθ = opposite/adjacent = 1/-3 = -1/3
α= arcsin(1/√10) ≈ 18
θ =180 -18 ≈162
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-5x + 3x > 22
-2x > 22
x < 22/-2 (greater than sign changes to less than sign when you divide with negative number)
x < -11
Answer:
Step 2 is wrong.
It should be 3x – <u>18</u> + 4x + 12 – 6x
Explanation:
3(x – 6) + (4x + 12) – 6x →
3x – 18 + 4x + 12 – 6x →
(3x + 4x – 6x) + (-18 + 12) →
x – 6