Let q be the number of 25 cent coins.
Let d be the number of 10 cent coins.
0.25q+0.10d= 3.95...(1)
q-d=6...(2)
(2)-> q-d= 6
q-d+d= 6+d
q= 6+d...(2a)
(2a)-> (1) 0.25q+0.10d=3.95
0.25(6+d)=0.10d= 3.95
1.95+0.25d+0.10d= 3.95
0.35d= 3.95-1.5
0.35d/0.35= 2.45/0.35
d= 7...(3)
(3)->(2) q-d= 6
q-7= 6
q=6+7
q= 13
There are 13 quarters and 7 dimes.
The answer should be -5x + 7y = -108
The answer to this would be 4(w-4)
:<span> </span><span>You need to know the derivative of the sqrt function. Remember that sqrt(x) = x^(1/2), and that (d x^a)/(dx) = a x^(a-1). So (d sqrt(x))/(dx) = (d x^(1/2))/(dx) = (1/2) x^((1/2)-1) = (1/2) x^(-1/2) = 1/(2 x^(1/2)) = 1/(2 sqrt(x)).
There is a subtle shift in meaning in the use of t. If you say "after t seconds", t is a dimensionless quantity, such as 169. Also in the formula V = 4 sqrt(t) cm3, t is apparently dimensionless. But if you say "t = 169 seconds", t has dimension time, measured in the unit of seconds, and also expressing speed of change of V as (dV)/(dt) presupposes that t has dimension time. But you can't mix formulas in which t is dimensionless with formulas in which t is dimensioned.
Below I treat t as being dimensionless. So where t is supposed to stand for time I write "t seconds" instead of just "t".
Then (dV)/(d(t seconds)) = (d 4 sqrt(t))/(dt) cm3/s = 4 (d sqrt(t))/(dt) cm3/s = 4 / (2 sqrt(t)) cm3/s = 2 / (sqrt(t)) cm3/s.
Plugging in t = 169 gives 2/13 cm3/s.</span>
