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3 years ago

15

Javier's fuel tank holds 12 3/4 gallons of gasoline when completely full. He had some gas in the tank and added 10.3 gallons of

gasoline to fill it completely.

1 answer:

GenaCL600 [577]3 years ago

6 0

Answer:

2.45 gallons of gasoline were in the tank before.

Step-by-step explanation:

Consider the provided information.

The tank can hold $12\frac{3}{4}$ gallons of gasoline when completely full. He had some gas in the tank and added 10.3 gallons of gasoline to fill it completely.

We need to find: How many gallons of gasoline were in the tank before Javier added some?

For this, we need to subtract added gallons of gasoline from the capacity of the tank.

$12\frac{3}{4}-10.3=\frac{51}{4}-10.3$

$=12.75-10.3$

$=2.45$

Hence, 2.45 gallons of gasoline were in the tank before.

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<h2>The missing measures of the angles are:</h2>

$m\angle 1= 60^{\circ}\\\\m\angle 2= 39^{\circ}\\\\m\angle 3= 21^{\circ}\\\\m\angle 4 = 39^{\circ}\\\\m\angle 5 = 21^{\circ}$

<u>Given</u>:

$m\angle Z = 138^{\circ}$

<em><u>Note the following:</u></em>

An equilateral triangle has all its three angles equal to each other. Each angle = $60^{\circ}$ .

This implies that, since $\triangle WXY$ is equilateral, therefore, $m\angle Y = m\angle XWY = m\angle XYW = 60^{\circ}$

Base angles of an isosceles triangle are congruent to each other.

This implies that, since $\triangle WZY$ is isosceles, therefore, $m\angle 3 = \angle 5$

Applying the above stated, let's find the measure of each angle:

Find $m\angle 1$

$m\angle 1 = 60^{\circ}$ (an angle in an equilateral triangle equals 60 degrees)

Find $m\angle 2$

$m\angle 2 = 60 - m \angle 3$

$m\angle 2 = 60 -\frac{1}{2}(180 - 138)$ $(Note: \frac{1}{2}(180 - 138) = 1 $ base $ angle $ of $ \triangle WZY)$

$m\angle 2 = 60 -21\\\\m\angle 2 = 39^{\circ}$

Find $m\angle 3$ and $m\angle 5$ (base angles of isosceles triangle WZY)

$m\angle 3 = \frac{1}{2}(180 - 138) (1 $ base $ angle $ of $ \triangle WZY)$

$m\angle 3 = \frac{1}{2}(42) \\\\m\angle3 = 21^{\circ}$

$m\angle3 = m\angle 5$ (base angles of isosceles triangle are congruent)

Therefore,

$m\angle5 = 21^{\circ}$

Find $m\angle 4$

$m\angle 4 = 60 - m\angle 5$

Substitute

$m\angle 4 = 60 - 21\\\\m\angle 4 = 39^{\circ}$

The missing measures of the angles are:

$m\angle 1= 60^{\circ}\\\\ m\angle 2= 39^{\circ}\\\\m\angle 3= 21^{\circ}\\\\m\angle 4 = 39^{\circ}\\\\m\angle 5 = 21^{\circ}$

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