Communitive Property of Addition
Answer:
4x and -2x
Step-by-step explanation:
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Answer:
Find the answers below
Step-by-step explanation:
Using m<X as the reference angle
Opposite YZ = 7
Adjacent XY = 10
Hypotenuse XZ = √149
Using the SOH CAH TOA identity
sinX = opp/hyp
sinX =YZ/XZ
sinX = 7/√149
For cos X
cos X = adj/hyp
cos X =10/√149
Using m<Z as reference angle;
Opposite XY = 10
Adjacent YZ = 7
Hypotenuse XZ = √149
Using the SOH CAH TOA identity
sinZ = opp/hyp
sinZ =10/√149
sinZ = 7/√149
For cos Z
cosZ = 7/√149
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.
<h3>
Answer: C) -1/2</h3>
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Explanation:
We'll apply the derivative to get
g(x) = 4x^3 + 3x^2
g ' (x) = 3*4x^2 + 2*3x
g ' (x) = 12x^2 + 6x
Set that equal to 0 and solve for x. We do this to find where the horizontal tangents are located.
g ' (x) = 0
12x^2 + 6x = 0
6x(2x + 1) = 0
6x = 0 or 2x+1 = 0
x = 0 or x = -1/2
Adding those two said solutions yields us the final answer choice C
