10.) you would take 5 and plug it in for b, then the problem would be 4 x 5 - 1
Following the rules of pemdas
(parentheses, exponents, multiply, divide, add, subtract)
you would do
4 x 5 = 20
20-1 = 19
11.) you would plug in 6 for m, therefore the problem will then be 5 + 2 x 6 squared
To solve it you would do 6 squared first and get 36
Then 2 x 36 is 72
5+72 is 77 which is the final answer
12.) following the other problems above you would plug in the number for the variable
In this problem you would plug in 1 for y squared so the problem would be
3 x 1 squared - 2
You would do 1 squared first and get 1
Then you would do 3 x 1 and get 3
And 3 - 2 is 1
13.) plug in 3 for a so you would get the problem
5 x 2 x 3 cubed
You would do 3 cubed first and get 27
5 x 2 is 10
10 x 27 is 270
14.) here you would plug in 5 for both c's so the problem would be
4 x 5 squared - 2 x 5
You would square the 5 first and get 25
4 x 25 is 100,
2 x 5 is 10
100-10 is 90
16.) you still would plug in so when you do you should get
40 - (32/4)
First you would do 32 and divide it by 4 and get 8
Then 40-8 is 32
I'll include all the word here at the bottom without words to maybe help you better
10.) 4b - 1 ; b = 5
4 x 5 - 1
20 - 1
19
11.) 5+2m squared ; m = 6
5 + 2 x 6 squared
5 + 2 x 36
5 + 72
77
12.) 3y squared - 2 ; y = 1
3 x 1 squared - 2
3 x 1 - 2
3 - 2
1
13.) 5x2a cubed ; a = 3
5 x 2 x 3 cubed
5 x 2 x 27
10 x 27
270
14.) 4c squared - 2c ; c = 5
4 x 5 squared - 2 x 5
4 x 25 - 2 x 5
100 - 2 x 5
100 - 10
90
16.) 40 - (32/r) ; r = 4
40 - (32/4)
40 - 8
32
