The student correctly: a. Estimates 10.357 g to 10.4 g, 12.062 g to 12.1 g, and 7.506 g as 7.5 g; finds the sum 30 g; shows work or model. b. Finds the sum 29.925 g and the difference 0.075 g. c. Finds the quotient 5.985 g and explains accurately the strategy used. d. Rounds 5.985 g to 6 g. And can explain using words WHY they solved the problem with certain strategies.

Step-by-step explanation:

A. Student estimates

10.357 to 10.4

12.062 to 12.1

7.506 to 7.1

Therefore, sum = 10.4 + 12.1 + 7.1

= 29.6 approximately it becomes 30 because we round to the nearest tens.

B. Sum without approximation gives

10.357 + 12.062 + 7.506

= 29.925

Therefore, difference becomes

Approximated value - Unapproximated value

30 - 29.925

= 0.075

2x+18+5x-9

2x+5x18-9

7x+9

7. 7

x= 1.4

hope it helps