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MArishka [77]
2 years ago
6

A cylindrical water tank near the town library is 20 meters high and has a circumference of 39 meters. What is the approximate v

olume, to the nearest whole number, of the water tank? Use 3.14 as an approximation for π and round to the nearest whole number.
Mathematics
1 answer:
nadezda [96]2 years ago
3 0

Step-by-step explanation:

height of the cylindrical tank = 20m

circumference of the tank = 39m

volumn of tank =  

perimetre formula = 2πrh

= 2 x 3.14 x r x 20 = 39

= 188.4r = 39

r = 188.4/39

r = 4.83

volumn =  πr^/2

h

= 3.14 x 4.83^2 x 2 x 20

= 1465.05

volumn of tank = 1465.05

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bekas [8.4K]

Answer:

B. 13

Step-by-step explanation:

Solve this by setting up system of equations:

0.25x + 0.10y = 3.95

x + y = 20

x equals the number of quarters and y equals the number of dimes. 0.25 is the value of a quarter and 0.10 is the value of a dime.

1. Multiply one equation to have the same coefficient as the other

0.10 · (x + y = 20) = 0.10x + 0.10y = 2

2. Subtract to find the value of one variable

 0.25x + 0.10y = 3.95

<u>- 0.10x + 0.10y = 2</u>

0.15x = 1.95

3. Solve for x by dividing both sides by 0.15

x = 13

3 0
3 years ago
4y + 12 = 7 - 5y thank you :)
riadik2000 [5.3K]
Y= -5 i am pretty sure
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3 years ago
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Step-by-step explanation:

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6 0
3 years ago
Mrs. McDonnell is making 25 paper cones to fill
viktelen [127]

Answer:

(A)29 cubic inch

Step-by-step explanation:

Diameter of the Cone =4 Inches

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Volume of a Cone =\frac{1}{3}\pi r^2h

First, we determine the radius, r.

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5 0
3 years ago
A large tank is filled to capacity with 600 gallons of pure water. brine containing 4 pounds of salt per gallon is pumped into t
nignag [31]
If A(t) is the amount of salt in the tank at time t, then the rate at which the amount of salt in the tank changes is given by

\dfrac{\mathrm dA(t)}{\mathrm dt}=\dfrac{4\text{ lbs}}{1\text{ gal}}\dfrac{6\text{ gal}}{1\text{ min}}-\dfrac{A(t)\text{ lbs}}{600\text{ gal}}\dfrac{6\text{ gal}}{1\text{ min}}
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Let's drop the units for now. We have

\dfrac{\mathrm dA(t)}{\mathrm dt}+\dfrac{A(t)}{100}=24
e^{t/100}\dfrac{\mathrm dA(t)}{\mathrm dt}+e^{t/100}\dfrac{A(t)}{100}=24e^{t/100}
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e^{t/100}A(t)=2400e^{t/100}+C
A(t)=2400+Ce^{-t/100}

We're given that the water is pure at the start, so A(0)=0, giving

A(0)=0=2400+Ce^{-0/100}\implies C=-2400

So the amount of salt in the tank (in lbs) at time t is

A(t)=2400\left(1-e^{-t/100}\right)
4 0
3 years ago
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