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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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By definition, the average rate of change is given by:

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Then, the AVR is:

AVR = \frac{18-(-2)}{3-(-2)}

AVR = \frac{18+2}{3+2}

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For f (x) = 3x - 8:

Evaluating for x =4:

f (4) = 3 (4) - 8\\f (4) = 12 - 8\\f (4) = 4

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Evaluating for x = 1:

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AVR = \frac{-4-(-4)}{1-(-1)}

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Answer:

from the greatest to the least value based on the average rate of change in the specified interval:


f(x) = x^2 + 3x interval: [-2, 3]

f(x) = 3x - 8 interval: [4, 5]

f(x) = x^2 - 5 interval: [-1, 1]

f(x) = x^2 - 2x interval: [-3, 4]


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Answer:

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Step-by-step explanation:

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The diagram is given below :

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Now, by using properties of equilateral triangle, median is perpendicular bisector and each angle is of 60°.

We get, ∠AMB = 90°. So, by linear pair ∠AMB + ∠PMB = 180° ⇒ ∠PMB = 90°. Also, ∠ABC = 60° and ∠ABP = 90° (given) So, ∠PBM = 30°

Since, AM is perpendicular bisector of BC. So,

MB = \frac{a}{2}

Now in ΔAMB , By using Pythagoras theorem

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Now, in ΔBMP :

sin\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}\\\\sin\thinspace 30^{o}=\frac{\text{MB}}{\text{PB}}\\\\PB=\frac{\text{MB}}{\text{sin 30}}\\\\PB=\frac{\frac{a}{2}}{\frac{1}{2}}\implies PB = a\\\\tan\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Base}}\\\\tan\thinspace 30^{o}=\frac{\text{MB}}{\text{PM}}\\\\PM=\frac{\text{MB}}{\text{tan 30}}\\\\PM=\frac{\frac{a}{2}}{\frac{1}{\sqrt3}}\implies PM=b= \frac{\sqrt{3}\cdot a}{2}

Perimeter of ABM = AB + PB + PM + AM

\text{Perimeter = }a+a+b+ \frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a + \frac{\sqrt{3}\cdot a}{2} +\frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a +\sqrt{3}\cdot a\\\\=(2+\sqrt3})\cdot a

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