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MAVERICK [17]
2 years ago
13

Five days a week, you carpool with 3 co-workers and take turns driving each week. It is 14 miles from your home to your office.

When you drive the carpool you must go an extra 4 miles to pick up your co-workers. If your car averages 18 miles per gallon, about how many gallons of gas should you save every 4 weeks by carpooling?
Mathematics
2 answers:
Vladimir [108]2 years ago
7 0

you will save 10 gallons

Sphinxa [80]2 years ago
4 0

Answer:

You will save 21.11 gallons of gas every 4 weeks by carpooling.

Step-by-step explanation:

It is 14 miles from your home to your office. Means 28 miles total while returning.

And if there is no carpool, you will drive =28\times5\times4=560 miles

And amount of gas used will be = \frac{560}{18}=31.11 gallons

-------------------------------------------------------------------------------------------------

Now, when you are car pooling, you will save the cost for three weeks means you will take out your car for 1 week only.

But you have to go additional 8 miles (including dropping off the co worker)

So, the total journey for 1 day will be of 14+14+4+4=36 miles

Hence, for 1 week, you will drive : 36\times5\times1=180 miles.

Gallons of gas used = \frac{180}{18}=10 gallons

-------------------------------------------------------------------------------------------------

So, gallons of gas saved = 31.11-10=21.11 gallons

Hence, you will save 21.11 gallons of gas every 4 weeks by carpooling.

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On a coordinate plane, a cube root function goes through (negative 2.5, negative 3.5), crosses the y-axis at (0, negative 3), ha
mart [117]

Answer:

g(x) = \sqrt[3]{x-1} - 2

Step-by-step explanation:

We want to find h and k in:

g(x) = \sqrt[3]{x-h} + k

At the inflection point, the second derivative is equal to zero, so:

g'(x) = \frac{1}{3} (x-h)^{\frac{-2}{3}}

g''(x) = \frac{1}{3} \frac{-2}{3}(x-h)^{\frac{-5}{3}} = 0

Then x - h = 0.

Inflection point is located at (1, -2), replacing this x value we get:

1 - h = 0

h = 1

We know that the point (-2.5, -3.5) belongs to the function, so:

-3.5 = \sqrt[3]{-2.5-1} + k

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All data, used or not, are shown in the picture attached.

4 0
2 years ago
Investigate the following harvesting model both qualitatively and analytically. If a constant number h of fish are harvested fro
zalisa [80]

Answer:

a. The population does not become extinct in finite time.

Step-by-step explanation:

The model for the population of the fishery is

dP/dt = P(a-bP)-h, P(0) = P_0

If we rearrange and replace the constants we have:

\frac{dP}{P(7-P)-49/4} =dt\\\\-4 (\frac{dP}{4(P-7)P+49}) =dt\\\\-4 \frac{dP}{(2P-7)^2} =dt\\\\-4 \int\frac{dP}{(2P-7)^2} =\int dt\\\\-4(-\frac{1}{2(2P-7)})=t+C\\\\\frac{2}{2P-7}=t+C\\\\ t=0 \,\,\, P(0)=P_0\\\\\frac{2}{2P_0-7}=0+C\\\\C=\frac{2}{2P_0-7}

Now we can calculate if the population become 0 in any finite time

\frac{2}{2P-7}=t+\frac{2}{2P_0-7}\\\\\frac{2}{2*0-7}=t+\frac{2}{2P_0-7}\\\\-\frac{2}{7}=t+\frac{2}{2P_0-7}\\\\

To be a finite time, t>0

t=-\frac{2}{7}-\frac{2}{2P_0-7}=0\\\\-\frac{2}{2P_0-7}=\frac{2}{7}\\\\7-2P_0=7\\\\P_0=0

We can conclude that the only finite time in which P=0 is when the initial population is 0.

Because P0 is a positive constant, we can say that the population does not become extint in finite time.

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