X³ = 125/27
Cube root both sides to isolate the variable:
∛x³ = ∛(125/27)
x = ∛(125/27)
∛125 = 5, ∛27 = 3
x = 5/3
Answer:
Step-by-step explanation:
1,8
5,7
2,3
Basically you find the data points and fill in the values. For Point J it is (2,4). They want you to do this: (2-4,4-6)...it then becomes (-2,-2).
So
J: (-2,-2)
K: (-1, -4)
I: (-2, -5)
Those are your new points. I hope this helps love! :)
The general form of a polynomial is: x² + bx + c = 0. Substituting the roots,
3² + b(3) + c = 0
3b + c = -9 --> eqn 1
(5 + √5)² + (5 +√5)(b) + c = 0
(5+√5)b + c = -30-10√5 --> eqn 2
Solving both equations simultaneously by subtracting eqn 2 from eqn 1,
(-2 - √5)b = 21 + 10√5
b = -8 - √5
Using eqn 1,
3(-8 - √5) + c = -9
-24 - 3√5 + c = -9
c = -9 + 24 + 3√5
c = 15 + 3√5
Hence, the polynomial is: <em>f(x) = x² + (-8 - √5)x + (15 + 3√5)</em>.
Answer:
x = 8
z = 65
Step-by-step explanation:
Through various rules, we are able to assert: (6x + 67)° = 115°.
Now, we want to solve for x:
6x + 67 = 115
6x = 48
x = 8
We know that z° is equal to the complement of (6x + 67)°, and together these must sum to 180°. Therefore, we can simply take 180° - 115° to find that z° = 65°.
