Answer:
10 quarters = $2.50
10 nickels = $0.50
that leaves $0.20 for other coins (dimes / pennies)
Step-by-step explanation:
First, suppose she has only quarters and nickels and no other coins. Then if C is the identical number of coins of each type, then 5C + 25C = 320, so 30C = 320 and 3C = 32, but there is no integer solution to this. So she must have at least one other type of coin.
Assume she has only quarters, nickels, and dimes. Then if D is the number of dimes, 5C + 25C + 10D = 320, which means 30C + 10D = 320, or 3C + D = 32. The smallest D can be is 2, leaving 3C = 30 and thus C = 10. So in this scenario she would have 10 quarters, 10 nickels, and two dimes to make $2.50 + $0.50 + $0.20 = $3.20.
This has to be the highest number, because if she had 11 quarters and 11 nickels, that alone would add up to 11(0.25) + 11(0.05) = $3.30, which would already be too much.
Hopefully that helps I’m sry but I had to crop the image so it isn’t perfect
Complementary angles add up to 90.
Say the complementary angle is A.
Angle A + Angle B = 90
Angle A + 60 = 90
Angle A = 30 degrees
Hope this helps :)
Answer:
The objective function in terms of one number, x is
S(x) = 4x + (12/x)
The values of x and y that minimum the sum are √3 and 4√3 respectively.
Step-by-step explanation:
Two positive numbers, x and y
x × y = 12
xy = 12
S(x,y) = 4x + y
We plan to minimize the sum subject to the constraint (xy = 12)
We can make y the subject of formula in the constraint equation
y = (12/x)
Substituting into the objective function,
S(x,y) = 4x + y
S(x) = 4x + (12/x)
We can then find the minimum.
At minimum point, (dS/dx) = 0 and (d²S/dx²) > 0
(dS/dx) = 4 - (12/x²) = 0
4 - (12/x²) = 0
(12/x²) = 4
4x² = 12
x = √3
y = 12/√3 = 4√3
To just check if this point is truly a minimum
(d²S/dx²) = (24/x³) = (8/√3) > 0 (minimum point)
