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Ostrovityanka [42]
3 years ago
11

Two pipes p and Q fill a water tank. Each pipe fills the tank with water at a constant rate. Pipe P fills the tank with 20 gallo

ns of water in 4 minutes. The equation in equals 6t shows the number of gallons of water n with which pipe you fills the tank in t minutes. Which pipe fills the tank with more water each minute and by how much
Mathematics
1 answer:
Lady_Fox [76]3 years ago
4 0

Answer:

Pipe Q

Step-by-step explanation:

Two pipes p and Q fill a water tank. Each pipe fills the tank with water at a constant rate. Pipe P fills the tank with 20 gallons of water in 4 minutes. The equation in equals n = 6t shows the number of gallons of water n with which pipe Q fills the tank in t minutes. Which pipe fills the tank with more water each minute and by how much

Solution:

Pipe P fills the tank with 20 gallons of water in 4 minutes. If n represent the number of gallons of water that fills the tank in t minutes, then for pipe P:

n = 20 gallons / 4 minutes *t

n = 5t

Also, the number of gallons of water that fills the tank in t minutes for pipe Q is given by:

n = 6t

Since the rate at which pipe Q fills the tank (i.e. 6t) is greater than the rate at which pie P fills the tank (i.e. 5t), hence Pipe P  fills the tank with more water each minute by 6 gallons while pipe P fills by 5 gallons each minute.

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What’s the expression 1/8+2/5
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The answer is 21/40

Step-by-step explanation:

(1×5)+(2×8)8×5

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2 years ago
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7 0
3 years ago
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Rationalize the denominator and simplify 1/√5
lozanna [386]

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\frac{\sqrt{5} }{5}

Step-by-step explanation:

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8 0
3 years ago
The value of an investment A (in dollars) after t years is given by the function A(t) = A0ekt. If it takes 10 years for an inves
Ludmilka [50]

Answer:

20 years.

Step-by-step explanation:

We have been given a formula A(t)=A_0\cdot e^{kt}, which represents the  value of an investment A (in dollars) after t years.

Substitute the given values:

\$3,000=\$1,000\cdot e^{k*10}

Let us solve for k.

\frac{\$3,000}{\$1,000}=\frac{\$1,000\cdot e^{k*10}}{\$1,000}

3=e^{k*10}

Take natural log of both sides:

\text{ln}(3)=\text{ln}(e^{k*10})

Using property \text{ln}(a^b)=b\cdot \text{ln}(a), we will get:

\text{ln}(3)=10k\cdot\text{ln}(e)

We know that \text{ln}(e)=1, so

\text{ln}(3)=10k\cdot 1

\text{ln}(3)=10k

\frac{\text{ln}(3)}{10}=\frac{10k}{10}

\frac{\text{ln}(3)}{10}=k

\$9,000=\$1,000\cdot e^{\frac{\text{ln}(3)}{10}*t}

Dividing both sides by 1000, we will get:

9=e^{\frac{\text{ln}(3)}{10}*t}

Take natural log of both sides:

\text{ln}(9)=\text{ln}(e^{\frac{\text{ln}(3)}{10}*t)

\text{ln}(9)=\frac{\text{ln}(3)}{10}*t\cdot\text{ln}(e)

\text{ln}(9)=\frac{\text{ln}(3)}{10}*t\cdot1

10*\text{ln}(9)=10*\frac{\text{ln}(3)}{10}*t

10\text{ln}(9)=\text{ln}(3)*t

10\text{ln}(3^2)=\text{ln}(3)*t

2\cdot 10\text{ln}(3)=\text{ln}(3)*t

20\text{ln}(3)=\text{ln}(3)*t

Divide both sides by \text{ln}(3):

\frac{20\text{ln}(3)}{\text{ln}(3)}=\frac{\text{ln}(3)*t}{\text{ln}(3)}

20=t

Therefore, it will take 20 years for the investment to be $9,000.

5 0
3 years ago
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