Answer:
Just add the first 2 masses together and subtract with the 3rd mass.
I started by labeling the right angle (Angle C) 90º. Next, I wrote down everything in one equation.
2x + 90 + 3x - 20 = 180º (180 degrees in a triangle)
Next, I add 20 on both sides.
2x + 90 + 3x = 200º
I combine like terms (2x and 3x)
5x + 90 = 200º
I subtract 90 from both sides.
5x = 110º
Divide 110 by 5 to get x.
x = 22
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For problem two, I label all the angles I know.
49º + 80º + r = 180º
I add 80 and 49.
129º + r = 180º
I subtract 180 and 129 and get 51º, which is your angle for R.
For angle X, you know that angle R plus angle X equals half of a circle, which is 180º
We know from before that 129º is 180º without R, so X is 129º
I hope this helps! Let me know if I'm wrong!
I can’t see the question very well
Answer:
z = x^3 +1
Step-by-step explanation:
Noting the squared term, it makes sense to substitute for that term:
z = x^3 +1
gives ...
16z^2 -22z -3 = 0 . . . . the quadratic you want
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<em>Solutions derived from that substitution</em>
Factoring gives ...
16z^2 -24z +2z -3 = 0
8z(2z -3) +1(2z -3) = 0
(8z +1)(2z -3) = 0
z = -1/8 or 3/2
Then we can find x:
x^3 +1 = -1/8
x^3 = -9/8 . . . . . subtract 1
x = (-1/2)∛9 . . . . . one of the real solutions
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x^3 +1 = 3/2
x^3 = 1/2 = 4/8 . . . . . . subtract 1
x = (1/2)∛4 . . . . . . the other real solution
The complex solutions will be the two complex cube roots of -9/8 and the two complex cube roots of 1/2.