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
Answer:
Yes
Step-by-step explanation:
By rounding to the 10ths place, we can easily see that 6.3 is greater than 6.04. 6.3 is already rounded to the 10ths place, but 6.04 rounded to the 10ths place is 6.0.
6.3 is clearly more than 6.0, therefore 6.3 is greater than 6.04.
Answer: 2
Step-by-step explanation:
(x1,y1) = (-1,0)
(x2,y2) = (3,8)
m = (8–0)/(3-(-1))
m = 8/(3+1)
m = 8/4
m = 2
