this function would count the profit of a month. To get the profit of longer periods you would need to do something like this:
[text]g(t) = t * f(n)[/text] where t is time.
I constructed the first function by simply looking at how much would one dog give profit. Since the fee is 500 and 200 of it is going to be spent away, 300 is left. So, profit from one dog is 300 dollars. However, if they don't have enough dogs, the doggy daycare company would get in debt because of 2000 dollars that it needs a month to operate, that's why I subtracted 2000 from the profit.
I just met a friend for the first year old man who
<h3>
Answer: Choice A</h3>
0.10d + 0.05n = 2.25
d = 2n
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Explanation:
n = number of nickels
d = number of dimes
we have twice as many dimes as nickels, so we could have something like 18 dimes and 9 nickels. Algebraically we would write d = 18 and n = 9. The equation that fits this is d = 2n. The answer is between A and B at this point.
We rule out choice B since the 0.05d is incorrect. It should be 0.10d to represent the total value of all the dimes.
Choice A is correct
d = number of dimes
0.10d = value of all the dimes in dollars
n = number of nickels
0.05n = value of all the nickels in dollars
0.10d + 0.05n = total value from both coins combined
2.25 = total value from both coins combined
0.10d + 0.05n = 2.25
Answer: $6.46
1.40 + .40 + 1.68 + 1.14 + 1.84 = 6.46
(-7) negative 7 i had some of these questions in pre-algrebra
JD
