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2 years ago

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For a client, a landscaper purchases 3 oak trees and 4 maple trees for $380. For his own home, the landscaper purchases 2 oak tr

ees and 5 maple trees for $370. Which augmented matrix represents the situation?

1 answer:

Pachacha [2.7K]2 years ago

5 0

Answer:

attached below

Step-by-step explanation:

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What is the correlation coefficient of the linear fit of the data shown below, to the nearest hundredth?

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According to one company’s profit model, the company has a profit of 0 when 10 units are sold and a maximum profit of $18,050 wh

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Is the line through points P(-8,-10) and Q(-5,-12) perpendicular to the line through points R(9,-6) and S(17,-5)? Explain.

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Consider this option:
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2 years ago

A certain cylindrical tank holds 20,000 gallons of water which can be drained from the bottom of the tank in 20 minutes the volu

vazorg [7]

Answer:

Impossible. t=30 minutes.

Step-by-step explanation:

We have the function:

$v(t)=20000(1-\frac{t}{20})^2$

Where v(t) represents the amount of gallons remaining after t minutes.

We want to find at what time t is the <em>instanteous </em>rate of change from the tank 1000 gallons per minute.

In order to determine the instantaneous rate of change, let's find the derivative of our function. So, take the derivative of both sides with respect to our time t:

$\frac{d}{dt}[v(t)]=\frac{d}{dt}(20000(1-\frac{t}{20})^2]$

On the right, let's move the coefficient outside:

$v'(t)=20000\frac{d}{dt}[(1-\frac{t}{20})^2]$

To differentiate, let's use the chain rule, which is:

$u(v(x))=u'(v(x))\cdot v'(x)$

Our u(x) is x² and v(x) is (1-t/20). So, u'(x) is 2x and v'(x) is -1/20. Therefore:

$v'(t)=20000(2(1-\frac{t}{20})\cdot -\frac{1}{20})$

Simplify:

$v'(t)=-1000(2(1-\frac{t}{20}))$

Simplify:

$v'(t)=-1000(2-\frac{t}{10})$

Distribute:

$v'(t)=100t-2000$

So, the instantaneous rate of change after time t is given by the above function.

To find when the instantaneous rate of change of the water is 1000 gallons per minute, substitute 100 for v'(t) and solve for t. So:

$1000=100t-2000$

Solve for t. Add 2000 to both sides:

$3000=100t$

Divide both sides by 100:

$t=30$

So, after 30 minutes, the instantaneous rate of change will be 1000 gallons per minute.

However, if we go back to our original function, our domain t is only defined between 0 and 20 minutes.

So, it is impossible for our instantaneous rate of change to ever reach 1000 gallons per minute.

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2 years ago