A jar of quarters and dimes contains $20.00. There are 110 coins in all. How many of each are there?

Mathematics

1 answer:

VikaD [51]3 years ago

6 0

Answer:

There are 50 dimes and 110-50 = 60 quarters in the collection.

Step-by-step explanation:

Let D be the number of dimes.

Then the number of quarters is (110-D).

The "value equation" is

10D + 25*(110-D) = 2000.

Simplify and solve for D:

10D + 2750 - 25D = 2000,

-15D = 2000 - 2750,

-15D = -750,

D = %28-750%29%2F%28-15%29 = 50.

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Mashcka [7]

Answer:

a. After the first bounce, the ball will be at 85% of 8 ft. After 2 bounces, it'll be at 85% of 85% of 8 feet. After 3 bounces, it'll be at (85% of) (85% of) (85% of 8 feet). You can see where this is going. After n bounces the ball will be at

$h(n) = (85\%)^n 6ft = (\frac{17}{20})^n(6ft\times 12\frac{in}{ft}) = (\frac{17}{20})^n\times72'$

b. After 8 bounces we can apply the previous formula with n = 8 to get

$h(8)=(\frac{17}{20})^8\times 72' = 19.62'$

c. The solution to this point requires using exponential and logarithm equations; a more basic way would be trial and error using the previous $h(n)$ increasing the value of n until we find a good value. I recommend using a spreadsheet for that; the condition will lead to the following inequality:

$1>(\frac{17}{20})^n\times 72$ Let's first isolate the fraction by dividing by 72.

$(\frac{17}{20})^n$ Now, to get numbers we can plug in a calculator, let's take the natural logarithm of both sides:

$ln ((\frac{17}{20})^n) < ln \frac 1{72} \rightarrow n\times ln (\frac{17}{20}) < -ln72$ . Now the two quantities are known - or easy to get with any calculator, replacing them and solving for n we get:

$-(0.16)n < - 4.27 \rightarrow n>26.31$ Now, since n is an integer - you can't have a fraction of a bounce after all, you pick the integer right after that, or n>27.