X actually equals the cube root of 6
![{x}^{3} - 6 = 0 \\ {x}^{3} = 6 \\ x = \sqrt[3]{6}](https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B3%7D%20%20%20-%206%20%3D%200%20%5C%5C%20%20%7Bx%7D%5E%7B3%7D%20%20%3D%206%20%5C%5C%20x%20%3D%20%20%5Csqrt%5B3%5D%7B6%7D%20)
Answer:
a) for all values of x that are in the domains of f and g.
b) for all values of x that are in the domains of f and g.
c) for all values of x that are in the domains of f and g with g(x)≠0
Step-by-step explanation:
a) By definition (f+g)(x)=f(x)+g(x). Then x must be in the domain of f and g.
b) By definition (fg)(x)=f(x)g(x). Then x must be in the domain of f and g.
c) By definition (f/g)(x)=f(x)/g(x). Then x must be in the domain of f and g and g(x) must be different of 0.
Answer:
15
Step-by-step explanation:
30 subtract 15 = 15 and both the 10 and the other 10 are the same
Answer:
option A
Step-by-step explanation:
We can find the standard form by expanding the equation:
y = 3(x-4)^2 - 22
y = 3(x^2 - 8x + 16) - 22
y = 3x^2 - 24x + 48 - 22
y= 3x^2 - 24x + 26
So the correct option is option A
That would be A :D It would be A because you would do keep change flip. You would keep 5/6, change the division sign to multiplication, then flip the 5/16 to 16/5 (the reciprocal)
