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Art [367]
3 years ago
5

Dylan has a bank that sorts coins as they are dropped into it. A panel on the front displays the total number of coins inside as

well as the total value of these coins. The panel shows 90 coins with a value of $17.55 inside of the bank.
If Dylan only collects dimes and quarters, write a system of equations in two variables or an equation in one variable that could be used to model this situation.




Using your equation or system of equations, algebraically determine the number of quarters and the number of dimes Dylan has in his bank
Mathematics
1 answer:
vesna_86 [32]3 years ago
4 0

Answer:

1). System of equations- D + Q = 90 and D + 2.5Q = 175.50

2). There are 33 dimes and 57 quarters in the bank.

Step-by-step explanation:

Dylan has a bank that sorts coins as they are dropped in it.

A panel shows the total number of coins inside as well as the total value of these coins.

Let the number of dimes in the bank = D

and the number of quarters in the bank = Q

If the panel shows total number of coins = 90

Then the equation will be

D + Q = 90 -------(1)

And the panel displays the amount of the coins = $17.55

Then equation will be

0.10D + 0.25Q = 17.55 [1 Dime = $0.10 and 1 quarter = $0.25]

D + 2.5Q = 175.50 ------------(2)

Now we subtract equation (1) form equation (2)

D + 2.5Q - (D + Q) = 175.50 - 90

D + 2.5Q - D - Q = 85.5

1.5Q = 85.5

Q = \frac{85.5}{1.5}

   = 57

By putting Q = 57 in the equation (1)

D + 57 = 90

D = 90 - 57 = 33

Therefore, there are 33 dimes and 57 quarters in the bank.

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The function f(x) is a cubic function and the zeros of f(x) are -3, -2, and 1. The y-intercept of f(x) is -24. Write the equatio
Serga [27]

Answer:

f(x) = 4x³+16x²+4x-24

Step-by-step explanation:

Factors are (x+3)(x+2)(x-1)

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(x+3)(x²+x-2)

x³+x²-2x+3x²+3x-6

x³+4x²+x-6

This could have been the function but the intetcept is 6 here.

So we'll take a miltiple of this one such that the intercept becomes-24

-24/-6 = 4

That multiple is 4

f(x) = 4(x³+4x²+x-6)

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3 years ago
What are the first, second, and third frequencies audible from a 20 cm long organ pipe when (A) only one end is open and when (B
kotykmax [81]
A) The answers are:
the first frequency - 428.75 Hz
the second frequency - 1286.25 Hz
the third frequency - 2143.75 Hz

The frequency (when the pipe is closed) is: f = v(2n - 1)/4L
v - the speed of sound
n - the frequency order
L - the length of the organ pipe

We know:
v = 343 m/s
L = 20 cm = 0.2 m

1. The first frequency (n = 1):
f = 343 * (2 * 1 - 1) / 4 * 0.2 = 343 * 1 / 0.8 = 428.75 Hz


2. The second frequency (n = 2):
f = 343 * (2 * 2 - 1) / 4 * 0.2 = 343 * 3 / 0.8 = 1286.25 Hz


3. The third frequency (n = 3):
f = 343 * (2 * 3 - 1) / 4 * 0.2 = 343 * 5 / 0.8 = 2143.75 Hz


B) The answers are:

the first frequency - 857.5 Hz
the second frequency - 1715 Hz
the third frequency - 2572.5 Hz

The frequency (when the pipe is open) is: f = vn/2L
v - the speed of sound
n - the frequency order
L - the length of the organ pipe

We know:
v = 343 m/s
L = 20 cm = 0.2 m

1. The first frequency (n = 1):
f = 343 * 1 / 2 * 0.2 = 343 / 0.4 = 857.5 Hz


2. The second frequency (n = 2):
f = 343 * 2 / 2 * 0.2 = 686 / 0.4 = 1715 Hz


3. The third frequency (n = 3):
f = 343 * 3 / 2 * 0.2 = 1029 / 0.4 = 2572.5 Hz
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