Greatest Common Factor - GCF
Factors of 24: 1; 2; 3; 4; 6; 8; 12; 24
Factors of 60: 1; 2; 3; 4; 5; 6; 10; 12; 15; 20; 30; 60
Factors of 72: 1; 2; 3; 4; 6; 8; 9; 12; 18; 23; 24; 36; 72
<span>GCF(24; 60; 72) = 12
</span>other method
24|2 60|2 72|2
12|2 30|2 36|2
6|2 15|3 18|2
3|3 5|5 9|3
1| 1| 3|3
1|
GCF(24; 60; 72) = 2 · 2 · 3 = 12
Answer: 1500000
Step-by-step explanation:
The half of 3,000,000 is 1500000 so 1500000
Answer:
Step-by-step explanation:
Factoring 
results in
Answer:
n =-19
Step-by-step explanation:
3* (n+7) = -36
Distribute
3n+21 = -36
Subtract 21 from each side
3n+21-21 = -36-21
3n = -57
Divide by 3
3n/3 = -57/3
n =-19
<span>The fact that Helen’s indifference curves touch the axes should immediately make you want to check for a corner point solution. To see the corner point optimum algebraically, notice if there was an interior solution, the tangency condition implies (S + 10)/(C +10) = 3, or S = 3C + 20. Combining this with the budget constraint, 9C + 3S = 30, we find that the optimal number of CDs would be given by 3018â’=Cwhich implies a negative number of CDs. Since it’s impossible to purchase a negative amount of something, our assumption that there was an interior solution must be false. Instead, the optimum will consist of C = 0 and Helen spending all her income on sandwiches: S = 10. Graphically, the corner optimum is reflected in the fact that the slope of the budget line is steeper than that of the indifference curve, even when C = 0. Specifically, note that at (C, S) = (0, 10) we have P C / P S = 3 > MRS C,S = 2. Thus, even at the corner point, the marginal utility per dollar spent on CDs is lower than on sandwiches. However, since she is already at a corner point with C = 0, she cannot give up any more CDs. Therefore the best Helen can do is to spend all her income on sandwiches: ( C , S ) = (0, 10). [Note: At the other corner with S = 0 and C = 3.3, P C / P S = 3 > MRS C,S = 0.75. Thus, Helen would prefer to buy more sandwiches and less CDs, which is of course entirely feasible at this corner point. Thus the S = 0 corner cannot be an optimum]</span>
