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Mrac [35]
3 years ago
9

The diameter of a cylindrical propane gas tank is 6 feet. The total volume of the rank is 1017.9 cubic feet. Find the length of

the tank. (Round your answer to the nearest foot.)
Mathematics
1 answer:
Luden [163]3 years ago
7 0

These are the things that I found on Google

Here are the most common propane tanks:

→20 lb tank: 18” tall and 12” in diameter. Holds 5 gallons of propane. ...

→33 lb tank: 2 feet tall and 1 foot in diameter. Holds 8 gallons of propane. ...

→100 lb tank: 4 feet tall and 18” diameter. ...

→420 lb tank: 4 feet tall by 3 feet diameter. ...

→500 gallon tank: 5 feet tall by 10 feet long.

Hope this helps ^_^  <em>:3</em>

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Find the accumulated value of an investment of $20,000 for 7 years at an interest rate of 5.5% if the money is:
Amiraneli [1.4K]

Answer:

  • See below

Step-by-step explanation:

<u>Given:</u>

  • Investment P = $20000
  • Time t = 7 years
  • Interest rate r = 5.5% = 0.055

a. <u>compounded semiannually, n = 2</u>

  • A = P*(1 + r/n)^{nt} = 20000(1+0.055/2)^{2*7} = 29239.88

b. <u>compounded quarterly, n = 4</u>

  • A = P*(1 + r/n)^{nt} = 20000(1+0.055/4)^{4*7} =  29315.30

c. <u>compounded monthly, n = 12</u>

  • A = P*(1 + r/n)^{nt} = 20000(1+0.055/12)^{12*7} =  29366.44

d. <u>compounded continuously</u>

  • A = Pe^{rt}=20000*e^{0.055*7}=29392.29
5 0
3 years ago
Read 2 more answers
3(x-4)=2(-2+1)<br> I don’t know this equation
Valentin [98]

Answer:

x=6

Step-by-step explanation:

8 0
4 years ago
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Devon is buying 3 greeting cards that cost $0.97 each. How can rounding be used to estimate the total cost of the cards?
LuckyWell [14K]

Answer:

The second answer, Since each card is about $1, the total cost is approximately 3 × 1 =$3

3 0
3 years ago
Two different radioactive isotopes decay to 10% of their respective original amounts. Isotope A does this in 33 days, while isot
Andrews [41]

Answer:

The approximate difference in the half-lives of the isotopes is 66 days.

Step-by-step explanation:

The decay of an isotope is represented by the following differential equation:

\frac{dm}{dt} = -\frac{t}{\tau}

Where:

m - Current mass of the isotope, measured in kilograms.

t - Time, measured in days.

\tau - Time constant, measured in days.

The solution of the differential equation is:

m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }

Where m_{o} is the initial mass of the isotope, measure in kilograms.

Now, the time constant is cleared:

\ln \frac{m(t)}{m_{o}} = -\frac{t}{\tau}

\tau = -\frac{t}{\ln \frac{m(t)}{m_{o}} }

The half-life of a isotope (t_{1/2}) as a function of time constant is:

t_{1/2} = \tau \cdot \ln2

t_{1/2} = -\left(\frac{t}{\ln\frac{m(t)}{m_{o}} }\right) \cdot \ln 2

The half-life difference between isotope B and isotope A is:

\Delta t_{1/2} = \left| -\left(\frac{t_{A}}{\ln \frac{m_{A}(t)}{m_{o,A}} } \right)\cdot \ln 2+\left(\frac{t_{B}}{\ln \frac{m_{B}(t)}{m_{o,B}} } \right)\cdot \ln 2\right|

If \frac{m_{A}(t)}{m_{o,A}} = \frac{m_{B}(t)}{m_{o,B}} = 0.9, t_{A} = 33\,days and t_{B} = 43\,days, the difference in the half-lives of the isotopes is:

\Delta t_{1/2} = \left|-\left(\frac{33\,days}{\ln 0.90} \right)\cdot \ln 2 + \left(\frac{43\,days}{\ln 0.90} \right)\cdot \ln 2\right|

\Delta t_{1/2} \approx 65.788\,days

The approximate difference in the half-lives of the isotopes is 66 days.

4 0
3 years ago
Read 2 more answers
Help please, find the blanks
Aleks [24]

Answer:

OQ = 16

PR = 32

m<QRS = 90 degrees

Step-by-step explanation:

Since we know that PQRS is a rectangle, we can use its properties to find that:

OQ, OS, OR, and OP are equivalent, meaning that they all equal 16. Therefore, OQ = 16.

PR = OR + OP, meaning that 16 + 16 = PR, meaning that PR = 32.

And finally, m<QRS = 90 degrees because all corner angles in a rectangle are right angles.

I hope this helps!

4 0
2 years ago
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