<u>ANSWER</u>
1. 
2. 
<u>QUESTION 1</u>
The first sentence is
.
Recall that;

We simplify the left hand side by applying this property to get;
.
.
We now rewrite the right hand side too in an index form to obtain;

We now equate the exponents to get;
.

.
<u>QUESTION 2</u>
The second sentence is 
We simplify the left hand side first to get;


We now rewrite the left hand side too in index form to obtain;

We equate the exponents to get;

This implies that;

or

Answer:
1. 21
2. 896
3. 20.9
Step-by-step explanation:
Use BODMAS/BIDMAS to solve.
You can use the distance equation.. d = rt
d = distance = 645
r = rate = 215 mph
t = time = ???? find the time
645 = 215t
645/215 = 215t / 215
3 = t
t = 3
It took 3 hours
For this question, we want to find out the greatest amount of area of pizza for the least amount of money:
So we have two pizzas with a 10 in. diameter. Let us take one of those pizzas.
The formula for area of a circle is

; So the area for one of these 10 in. pizzas would be

or 25

.
We have two of these pizzas, so the total area that we would get with $14 would be 25

*2 or 50

.
For the second case, we have one pizza with a 14 in. diameter.
The area of this pizza will be

or 49

.
So for $15 we can buy 49

worth of pizza.
The best deal would be to buy the two medium pizzas of 10 inch diameter, because for $14, you get 50

.
Answer:
Step-by-step explanation:
9/16=.5625
Round that to the nearest whole number would be 1