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
