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3 years ago

Tom has $5.05 in quarters and dimes. How many coins are quarters if the total

Mathematics

2 answers:

Oksanka [162]3 years ago

7 0

Answer:

15 quarters

Step-by-step explanation:

Assuming q = quarters and d = dimes...

First, set up the equations. q + d = 28 since we have 28 coins, and the latter is because a quarter is worth $0.25 and a dime is worth $0.1 - totaling $5.05.

q + d = 28

.25q + .10d = 5.05

Try to get one of the variables equal to the equation above. In this case, I tried to get the "q" variables equal to each other.

q + d = 28

4(.25q + .10d) = 4(5.05)

After this, you "subtract" the two equations from each other. Remember that d = 1d, and 1 - .4 = .6

q + d = 28

q + .4d = 20.20

Divide both sides by .6 to get d by itself.

.6d = 7.8

There are 13 dimes.

d = 13

Subtract 13 from the total coins (28) to get the amount of quarters.

28 - 13 = 15

13 dimes, 15 quarters. Hope this helped!

marissa [1.9K]3 years ago

6 0

Answer: The answer would be 7.03

Step-by-step explanation: cause you gotta get the 28 in there i hope im right

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Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 4cm and a height of 16cm, at the ra

bearhunter [10]

Answer:

$\frac{dh}{dt}=-\frac{1}{2\pi}cm/min$

Step-by-step explanation:

From similar triangles, see diagram in attachment

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$r=\frac{h}{4}$

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$V=\frac{1}{3}\pi (\frac{h}{4})^2h$

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$V=\frac{1}{48}\pi h^3$

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$\frac{dV}{dt}=\frac{\pi}{16}h^2 \frac{dh}{dt}$

We were given that water is drained out of the tank at a rate of $2cm^3/min$

This implies that $\frac{dV}{dt}=-2cm^3/min$ .

Since we want to determine the rate at which the depth of the water is changing at the instance when the water in the tank is 8cm deep, it means $h=8cm$ .