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Let us say that:
a = ones
b = fives
c = twenties
So that the total money is:
1 * a + 5 * b + 20 * c = 229
=> a + 5b + 20c = 229 -->
eqtn 1
We are also given that:
c = a – 5 -->
eqtn 2
a + b + c = 30 -->
eqtn 3
Rewriting eqtn 3 in terms of b:
b = 30 – a – c
Plugging in eqtn 2 into this:
b = 30 – a – (a – 5)
b = 35 – 2a -->
eqtn 4
Plugging in eqtn 2 and 4 into eqtn 1:
a + 5(35 – 2a) + 20(a – 5) = 229
a + 175 – 10a + 20a – 100 = 229
11a = 154
a = 14
So,
b = 35 – 2a = 7
c = a – 5 = 9
Therefore there are 14 ones, 7 fives, and 9 twenties.
Answer: Greater than
Answer: Greater than
Differentiate:x^2 + y^2 = 4
You get this:2x + 2yy' = 0
bring the x variable on the other side:2yy' = -2x
Now divide 2y to leave y' by itself:y' = -2x/2y
Answer: y' = -x/y
The given relation is not a implicit solution to the given differential equation.
