Answer:
anytime before or after 3 that day?
Step-by-step explanation:
◆ Define the variables:
Let the calorie content of Candy A = a
and the calorie content of Candy B = b
◆ Form the equations:
One bar of candy A and two bars of candy B have 774 calories. Thus:
a + 2b = 774
Two bars of candy A and one bar of candy B contains 786 calories
2a + b = 786
◆ Solve the equations:
From first equation,
a + 2b = 774
=> a = 774 - 2b
Put a in second equation
2×(774-2b) + b = 786
=> 2×774 - 2×2b + b = 786
=> 1548 - 4b + b = 786
=> -3b = 786 - 1548
=> -3b = -762
=> b = -762/(-3) = 254 calorie
◆ Find caloric content:
Caloric content of candy B = 254 calorie
Caloric content of candy A = a = 774 - 2b = 774 - 2×254 = 774 - 508 = 266 calorie
3 : 3,6,9,12,15,18,21,24,27,30
4 : 4,8,12,16,20,24,28,32,36
7: 7,14,21,28,35,42,49,56,63
9: 9,18,27,36,45,54,63,72,81
10%of 710 is 71
now we have 710-71=639
I honestly don’t know I’m just trying to get my question answered
