Let the number of nickels found be n, the number of dimes be d and the number of quarters be q.
i) "you have twice as many quarters as dimes and 42 coins in all."
means that 2d=q, and n+d+q=42
we can reduce the number of unknowns by substituting q with 2d:
n+d+q=42
n+d+2d=42
n+3d=42
We can write all n, d and q in terms of d as follows:
there are n=42-3d nickels, d dimes and q=2d quarters.
ii) In total there are $6.60 dollars,
1 nickel = 5 cent = $0.05
1 dime = 10 cent = $0.1
1 quarter = 25 cent = $0.25
thus
(42-3d)*0.05 + d*0.1 +2d*0.25= $6.60
2.1 - 0.15d+0.1d+0.5d=6.60
2.1+0.45d=6.6
0.45d=6.6-2.1=4.5
d=4.5/0.45=10
iii)
so, there are 10 dimes, 2d=2*10=20 quarters and 42-3d=42-3*10=12 nickels.
Answer: 10 dimes, 20 quarters, 12 nickels
14.4, if my calculator was correct
Four times a number (4 times x since x represents unknown) is at least (<) two (2)
4x>2
divide both sides by 2
2x>1
Let "a" and "b" be some number where:
a - b = 24
We want to find where a^2 + b^2 is a minimum. Instead of just logically figuring out that the answer is where a=b=12, I'll just use derivatives.
So we can first substitute for "a" where a = b+24
So we have (b+24)^2 + b^2 = b^2 +48b +576 + b^2
And that equals 2b^2 +48b +576
Then we take the derivative and set it equal to zero:
4b +48 = 0
4(b+12) = 0
b + 12 = 0
b = -12
Thus "a" must equal 12.
So:
a = 12
b = -12
And the sum of those two numbers squared is (12)^2 + (-12)^2 = 144 + 144 = 288.
The smallest sum is 288.
