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2 years ago

How many liters of gasoline to fill 13.2 gallon tank?

1 answer:

5 0

1. Find how many liters of gasoline to fill 13.2 gallon tank.
based on the metric conversion, there are 3.79 liters of liquid in every 1 liquid gallon.
Now, since we have 13.2 gallon tank, we need to multiply it with the given unit of conversion
=> 1 gallon = 3.79 liters
=> 13.2 gallon x 3.79 liters
=> 13.2 x 3.79
=> 50.028, therefore it needs around 50.028 liters of gasoline to be able to fill the 13.2 gallon tank

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Please help this test is a big part of my grade! (will give brainliest)

djyliett [7]

Answer:

Step-by-step explanation:

7 0

2 years ago

Andy has three planks of wood. Two planks are each 1.5 m long. One plank is 2.5m long. What is the total length of all the three

11Alexandr11 [23.1K]

Given:

Two planks are each 1.5 m long.

One plank is 2.5m long.

To find:

The total length of all the three planks.

Solution:

Two planks are each 1.5 m long. So,

First plank = 1.5 m

Second plank = 1.5 m

One plank is 2.5m long. So,

Third plank = 2.5 m

Now, the total length of all the three planks is

$\text{Total length}=\text{First plank + Second plank + Third plank}$

$\text{Total length}=1.5+1.5+2.5$

$\text{Total length}=5.5$

Therefore, the total length of all the three planks is 5.5 m.

4 0

3 years ago

A right triangle has a leg of 11 and a hypotenuse of 16. Find the missing leg (in simplest radical form)

igomit [66]

Answer:

11^2 + b^2 = 16^2

121 + b^2 = 256

b^2 = 135

b = 3 X q(15)

Step-by-step explanation:

8 0

3 years ago

What is the price of the items Nakita bought?

timurjin [86]

Answer:

B $10.45

Step-by-step explanation:

Fist you find how much the 2 boxes of cracker cost

$3.50 * 2 = $7

Then subtract the amount of 2 crackers by the $0.80 coupon

$7 - $0.80 = $6.20

Then add that amount to the cost of a jar of peanut butter

$6.20 + $4.85 = $11.05

Then add that about for with the amount of a package of juice boxes

$11.05 + $2.40 = $13.45

Then subtract the total amount to the $3.00 coupon

$13.45 - $3.00 = $10.45

B $10.45 is your answer

4 0

2 years ago

Vector u has a magnitude of 7 units and a direction angle of 330°. Vector v has magnitude of 8 units and a direction angle of 30

Dafna11 [192]

Keeping in mind that x = rcos(θ) and y = rsin(θ).

we know the magnitude "r" of U and V, as well as their angle θ, so let's get them in standard position form.

$\bf u=
\begin{cases}
x=7cos(330^o)\\
\qquad 7\cdot \frac{\sqrt{3}}{2}\\
\qquad \frac{7\sqrt{3}}{2}\\
y=7sin(330^o)\\
\qquad 7\cdot -\frac{1}{2}\\
\qquad -\frac{7}{2}
\end{cases}\qquad \qquad v=
\begin{cases}
x=8cos(30^o)\\
\qquad 8\cdot \frac{\sqrt{3}}{2}\\
\qquad \frac{8\sqrt{3}}{2}\\
y=8sin(30^o)\\
\qquad 8\cdot \frac{1}{2}\\
\qquad 4
\end{cases}$

$\bf u+v\implies \left( \frac{7\sqrt{3}}{2},-\frac{7}{2} \right)+\left( \frac{8\sqrt{3}}{2},4 \right)\implies \left( \frac{7\sqrt{3}}{2}+\frac{8\sqrt{3}}{2}~~,~~ -\frac{7}{2}+4\right)
\\\\\\
\left(\stackrel{a}{\frac{15\sqrt{3}}{2}}~~,~~ \stackrel{b}{\frac{1}{2}}\right)\\\\
-------------------------------$

$\bf tan(\theta )=\cfrac{b}{a}\implies tan(\theta )=\cfrac{\frac{1}{2}}{\frac{15\sqrt{3}}{2}}\implies tan(\theta )=\cfrac{1}{15\sqrt{3}}
\\\\\\
\measuredangle \theta =tan^{-1}\left( \cfrac{1}{15\sqrt{3}} \right)\implies \measuredangle \theta \approx 2.20422750397203^o$

8 0

2 years ago