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ehidna [41]
2 years ago
5

The Thomas family went for a Sunday drive. Before they left, Mr. Thomas noticed the gas tank was ¾ full. When they returned home

the gas tank was ⅓ full. If the gas tank holds 18 gallons, how many gallons of gas did the car use on the drive?
Mathematics
1 answer:
JulijaS [17]2 years ago
4 0

Answer:

7.5 gallons

Step-by-step explanation:

Given:

The Thomas family went for a Sunday drive.

Before they left, Mr. Thomas noticed the gas tank was ¾ full.

When they returned home the gas tank was ⅓ full.

Total capacity of the gas tank = 18 gallons

<u>Question asked:</u>

How many gallons of gas did the car use on the drive?

<u>Solu</u>tion:

Before they left, quantity of gas in the tank = \frac{3}{4} \times18=\frac{54}{4} =13.5\ gallons

When they returned, quantity of gas in the tank = \frac{1}{3} \times18=\frac{18}{3} =6\ gallons

Quantity of gas used on the drive = 13.5 - 6 = 7.5 gallons

Therefore, 7.5 gallons of gas used on the drive by Thomas family.

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masya89 [10]

We have to determine the complete factored form of the given polynomial

f(x)=6x^3-13x^2-4x+15.

Let x= -1 in the given polynomial.

So, f(-1)= 6(-1)^3-13(-1)^2-4(-1)+15 = -6-13+4+15 = 0

So, by factor theorem

(x+1) is a factor of the given polynomial.

So, dividing the given polynomial by (x+1), we get quotient as 6x^2-19x+15.

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3 years ago
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Answer:

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Perimiter: P=2a+2b

Minimum perimeter: [16,16]

Step-by-step explanation:

This is a problem of optimization with constraints.

We can define the rectangle with two sides of size "a" and two sides of size "b".

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To simplify and as we have only one constraint and two variables, we can express a in function of b as:

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The function we want to optimize is the diameter.

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To optimize we can derive the function and equal to zero.

dP/da=2+2\cdot (-1)\cdot\frac{256}{a^2}=0\\\\\frac{512}{a^2}=2\\\\a=\sqrt{512/2}= \sqrt {256} =16\\\\b=256/a=256/16=16

The minimum perimiter happens when both sides are of size 16 (a square).

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Answer:

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Step-by-step explanation:

37 increase 7% =

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6 0
3 years ago
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