Answer:
19. 1 do 4^5 which is 1024 then do 1024^0 which is 1
21. 0.000064 do 5^-2 which is 0.04 then do 0.04^3 which is 0.000064
23. 2.33 do 9^4 which is 6561 then do 6561^-2 which is 2.32305731 so you round it to 2.33
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<u>Answer:
</u>
Physliis invested 32000 dollar at 5% interest rate and 34000 dollar at 7% interest rate.
<u>Solution:</u>
Let Phyllis invest ‘x’ dollar at 5% per year and (66000-x) dollar at 7% per year.
We know,

In the question it is given that Simple interest earned from both the investments at the end of the year is $3980.
Using the given below equation, we will try to find out the investments at each rate.

x = 32000
We can calculate amount for 7% interest rate by,
(66000-32000) =34000
Thus Phyllis invested 32000 dollar at 5% interest rate and 34000 dollar at 7% interest rate.