(5b, -1z)
You have to multiply the B and the C by 5 to get the original equation or you could divide the original equation by 5
Answer:
A. x=2
b. x=28/3 and in decimal it is x=9.3 and in fraction it is
x= 9 1/3
c. x=20
d. x=20/11 in decamal x=1.81 in fraction it is x=1 9/11
Step-by-step explanation:
please give brainly
<h3>
Answer: 1</h3>
where x is nonzero
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Explanation:
We'll use two rules here
- (a^b)^c = a^(b*c) ... multiply exponents
- a^b*a^c = a^(b+c) ... add exponents
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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)
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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
Part A: Yes, the data represent a function. The definition of a function is a relation in which no value of x will have two different values of y.
(Every time you plug in 3 as x, you will always get 4 as y; it's ok if you plug in 3 and 5 as x and get the same y, you just can't get two different y's for one x; sorry, it is pretty confusing). None of the numbers in the table repeat, so we can safely say that the relation is a function.
Part B: All we have to do is plug in 11 for x in the function given to find the answer:

In the table, y = 8 when x = 11, but in the function given, y = 34 when x = 11, so the function given is greater.
Part C: To find the answer to C, just plug in 99 for f(x), as it tells you to do:
The <u>answer would be 6</u>. The number line which shows you the number 6 or any solution for 6 would be your answer.