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omeli [17]
4 years ago
8

A fuel tank contains 1 3/4 gallons of gasoline. Casey adds 33 1/3 gallons of

Mathematics
1 answer:
romanna [79]4 years ago
5 0

Answer:

The gallons of gas the tank contain now is <u>35 1/12</u>.

Step-by-step explanation:

Given:

Fuel tank contains 1 3/4 gallons of gasoline.

Casey adds 33 1/3 gallons of  gasoline to the tank.

Now, to find the total gallons of gas the tank contain.

Converting the mixed fractions into improper.

Gallons of gasoline in fuel tank =1\frac{3}{4}=\frac{7}{4}.

Gallons of gasoline Casey adds =33\frac{1}{3} =\frac{100}{3}.

<u>According to question:</u>

By adding we get the total gallons of gasoline now tank contain:

\frac{7}{4} +\frac{100}{3}

=\frac{7\times 3+100\times 4}{12}

=\frac{21+400}{12}

=\frac{421}{12}

=35\frac{1}{12}

Therefore, the gallons of gas the tank contain now is 35 1/12.

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Step-by-step explanation:

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3 years ago
The safety instructions for a 20 foot ladder say the ladder should not be inclined more than 70 degrees with the ground. suppose
Romashka [77]

Answer: the distance of the base of the house to the foot of the ladder is 6.84 feet

Step-by-step explanation:

The scenario is shown in the attached photo.

Right angle triangle ABC is formed when the ladder leans against the wall of the house.

AC = the height of the ladder

AB = x feet = distance of the base of the house to the foot of the ladder

BC is the wall of the building.

To determine x, we will apply trigonometric ratio

Cos # = adjacent/hypotenuse

Where

# = 70 degrees

Hypotenuse = 20

Adjacent = x

Cos 70 = x/20

x = 20cos70

x = 20 × 0.3420

x = 6.84 feets

4 0
3 years ago
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Nutka1998 [239]
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3 0
3 years ago
What is the distance from P to Q ?A. 7 unitsB. 5 unitsC. 1 unitD. 25 unitshow do I find the distance from P to Q?
Alex17521 [72]

to find the distance between 2 points we should apply the formula

d=\sqrt[]{(x_2-x_1)^2+(y_2-_{}y_1)^2_{}}

call point q as point 1 for reference in the formula and p as point 2

replace the coordinates in the formula

d=\sqrt[]{(3-(-1))^2+(-4-(-1))^2}

simplify the equation

\begin{gathered} d=\sqrt[]{(3+1)^2+(-4+1)^2} \\ d=\sqrt[]{4^2+(-3)^2} \\ d=\sqrt[]{16+9} \\ d=\sqrt[]{25} \\ d=5 \end{gathered}

the distance between the 2 points is 5 units

7 0
1 year ago
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