Answer:
When we have a function f(x) = y.
The rate of change between a and b (where a < b) is:
R = (f(b) - f(a))/(b - a)
In this case:
b = 3
a = 1
f(3) = 127
f(1) = 95
R = (127 - 95)/(3 - 1) = 16
The rate of change in that interval is 16.
No they are not equivalent
Weird way to write it but alright! (Sideways)
19pq^-2 x 5pq^6 = ?
These problems are pretty much single operations between each of the variables / constants.
So it's like this:
(19*5)(p*p)(q^-2*q^6) = ?
19*5 is 95.
For p*p remember that when two variables multiply there given powers add. In the case where the powers are not shown (like in the case of p*p) they are always assumed to be 1. So what is 1+1? 2.
p*p is p^2
For q^-2*q^6 it is the same deal with the previous problem. So now the problem looks like this:
-2 + 6 = 4
(The two is negative, because the power is negative 2)
So, q^4.
Our final answer is all of the combined.... like a so:
95p^2q^4
