Step-by-step explanation:
first box on the left
r=4
d=8
circumference= 2π*4= 8π
area = π*4*4= 16π
Second box on the left
d=6
r= 3
circumference= 2π*3= 6π
area =π*3*3= 9π
third box on the left
A=36π
A=36πarea= π*r*r
A=36πarea= π*r*rr= 6
A=36πarea= π*r*rr= 6d=12
A=36πarea= π*r*rr= 6d=12circumference= 2π*6= 12 π
the last box
C=18π
C=18πC= 2π*r
C=18πC= 2π*rr= 9
C=18πC= 2π*rr= 9d=18
C=18πC= 2π*rr= 9d=18area= π*9*9= 81π
The problem presents 2 variables and 2 conditions to follow to determine the approach in solving the problem. The variables are 52 cards, and 9 cards. The 2 conditions presented would be the teacher giving out one card to each student at a time to each student until all of them are gone. The second variable is more likely made as a clue and the important variable that gives away the approach to be used. The approach to be used is division. This is to ensure that there will be students receiving the 9 cards. Thus, we do it as this: 52 / 9 = ?
The answer would be 5.77778 (wherein 7 after the decimal point is infinite and 8 would just be the rounded of number). This would ensure us that there will be 5 students that can receive 9 cards but there will be 7 cards remaining which goes to the last student, which is supposed to be 8 since she gives one card to each student at a time to each student. So the correct answer would be just 4 students. The fifth student will only receive 8 cards and the last student would have 8, too.
