<h3>
Answer: 1</h3>
where x is nonzero
=======================================================
Explanation:
We'll use two rules here
- (a^b)^c = a^(b*c) ... multiply exponents
- a^b*a^c = a^(b+c) ... add exponents
------------------------------
The portion [ x^(a-b) ]^(a+b) would turn into x^[ (a-b)(a+b) ] after using the first rule shown above. That turns into x^(a^2 - b^2) after using the difference of squares rule.
Similarly, the second portion turns into x^(b^2-c^2) and the third part becomes x^(c^2-a^2)
-------------------------------
After applying rule 1 to each of the three pieces, we will have 3 bases of x with the exponents of (a^2-b^2), (b^2-c^2) and (c^2-a^2)
Add up those exponents (using rule 2 above) and we get
(a^2-b^2)+(b^2-c^2)+(c^2-a^2)
a^2-b^2+b^2-c^2+c^2-a^2
(a^2-a^2) + (-b^2+b^2) + (-c^2+c^2)
0a^2 + 0b^2 + 0c^2
0+0+0
0
All three exponents add to 0. As long as x is nonzero, then x^0 = 1
There is no picture or expression
Find the domain by finding where the function is defined. The range is the set of values that’s correspond with the domain
First you would plug in the 9 for x, then you would try to solve for y
3x^2 + 6x - 24
3(x^2 + 2x - 8)
3(x + 4)(x - 2) <==
