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:
A
Step-by-step explanation:
(5x4) - 6 / 2 = x
You can use the sum and difference identities which for cosine is cos(a+b)= cosacosb-sinasinb
Answer:

Step-by-step explanation:
we know that
The area of the figure is equal to the area of rectangle plus the area of two semicircles
<u>The area of rectangle is equal to</u>

<u>The area of the small semicircle is equal to</u>

-----> radius is half the diameter
substitute

<u>The area of the larger semicircle is equal to</u>

-----> radius is half the diameter
substitute

The area of the figure is equal to
