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alina1380 [7]
3 years ago
5

In the last race , the first car in the pit used 1 3/4 tanks of gas, the second car used 2 1/8 and the third car used 1 1/2 tank

s of gas. what was the total positive or negitive change in the amount of gas for all three cars
Mathematics
1 answer:
riadik2000 [5.3K]3 years ago
8 0

Answer: -5\frac{3}{8} tanks of gas


Step-by-step explanation:

Let the x be the number of tanks of gas, then

The amount of gas used by first car= 1\frac{3}{4}x=\frac{7}{4}x

The amount of gas used by second car= 2\frac{1}{8}x=\frac{17}{8}x

The amount of gas used by third car= 1\frac{1}{2}x=\frac{3}{2}x

Thus, the total change in the amount of gas for all three cars=\frac{7}{4}x+\frac{17}{8}x+\frac{3}{2}x

=x(\frac{7}{4}+\frac{17}{8}+\frac{3}{2})\\\\=x(\frac{14+17+12}{8})\\\\=x(\frac{43}{8})\\=5\frac{3}{8}x

Since, the amount of gas 5\frac{3}{8} tanks of gas is used for all three cars.

Therefore, the total amount would be negative .

hence, the answer is  -5\frac{3}{8} tanks of gas.

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4 0
3 years ago
Multiple Choice
Zanzabum

A) Answer: A. 22.5 cm, 30 cm, and 37.5 cm


Setting the unity measurement as x, the three sides will be 3·x, 4·x, and 5·x. The sum of the three sides will be (3 + 4 + 5)·x = 12·x; this corresponds to the perimeter, therefore:

12·x = 90 cm


We can solve for x:

x = 90 / 12 = 7.5 cm


Hence, the sides will be:

7.5 · 3 = 22.5 cm

7.5 · 4 = 30 cm

7.5 · 5 = 37.5 cm


This situation is represented in option A (which is also the only option whose sides add up to 90).



B) Answer: C. never true


We have:

6 + 8x - 9 = 11x + 14 - 3x


Bring all the terms with x on the left, by subtracting from both sides 11x and -3x, and all the numbers on the right by subtrcting from both sides 6 and -9:

6 + 8x - 9 - 11x - (-3x) - 6 - (-9) = 11x + 14 - 3x - 11x - (-3x) - 6 - (-9)


Cancel out the opposite terms from each side:

8x - 11x + 3x = 14 - 6 + 9


Combine likely terms:

0x = 17


Now you should ask yourself: what number multiplied by zero gives 17 as a result? The answer is none because a number multiplied by zero always equals zero.


Hence, the statement is never true.


C) Answer: r \leq \frac{1}{5}


We have:

5r + 4 ≤ 5


Subtract 4 from both sides:

5r + 4 - 4 ≤ 5 - 4


Combine likely terms:

5r ≤ 1


Divide both sides by 5:

\frac{5r}{5} \leq \frac{1}{5}

r \leq \frac{1}{5}


To graph the set of solutions, you need to set:

y₁ = 5x + 4 and

y₂ = +5


Plot these two lines:

y₂ is a horizontal line at a height of 5;

y₁ is a line passing through (0, 4) and (1, 9)


The set of solutions will be the part of the graph in which y₁ lies underneath y₂ (see picture attached), that happens for x \leq \frac{1}{5}.





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3 years ago
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What is a radius of a circle given by the equation x²-2x+8y-47=0
Zolol [24]
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8y=-x^2+2x+47
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Factor out coefficient.
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Convert the whole expression into square form.
y=-\frac{1}{8}\left(x-1\right)^2+\frac{1}{8}\cdot \:1+\frac{47}{8}
Simplify.
y=-\frac{1}{8}\left(x-1\right)^2+6
Subtract 6 from both sides.
y-6=-\frac{1}{8}\left(x-1\right)^2
Divide by coefficient.
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Rewrite in standard form.
4\left(-2\right)\left(y-6\right)=\left(x-1\right)^2

The radius of the circle will be -2. Also, no matter what the sign says, the radius has to always be positive. 
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Answer:

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