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:
793.25 mi/hr
Step-by-step explanation:
Given that:
The radius of the earth is = 3030 miles
The angular velocity = 
If a jet flies due west with the same angular velocity relative to the ground at the equinox;
We are to determine the How fast in miles per hour would the jet have to travel west at the 40th parallel for this to happen.
NOW;
Distance s is expressed by the relation
s = rθ
s = 793.25
The speed which depicts how fast in miles per hour the jet would have traveled is :
v = 793.25 mi/hr
Hence, the jet would have traveled 793.25 mi/hr due west at the 40th parallel for this to happen.
Answer:
because x is the middle integer of three consecutive integers
=> The remaining 2 numbers respectively are x - 1 and x + 1
=> the sum of these three integers is
x - 1 + x + x + 1 = 3x
<span>Joseph bought 23 laptop computers for school at a discount for $5,621. How much did he pay for each computer?</span>
