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daser333 [38]
4 years ago
9

Volume problem. Will give brainliest

Mathematics
1 answer:
Orlov [11]4 years ago
7 0

Answer:

<em>$ 33.6 to fill this tank, provided a community cost of $2.8 per gallon</em>

Step-by-step explanation:

1. Let us first find the volume of the gas the tank, by the general multiplication of Base * height ⇒ 11 inches * 1.25 feet * 1.75 feet. For the simplicity, we should convert feet ⇒ inches, as such: 1.25 feet = 1.25 * 12 inches = 15 inches, 1.75 feet = 1.75 * 12 inches = 21 inches. Now we have a common unit, let us find the volume ⇒ 11 in. * 15 in. * 21 in. = 3465 inches^3.

2. Let us say that the the average price of gas in my community is $2.8 per gallon. We would first have to convert inches ⇒ gallons provided 1 gallon = 231 inches: 3465/231 = 15 gallons.

4. Now simply multiply this price of 2.8 dollars per gallon by the number of gallons to receive the cost if the tank was full: 2.8 * 15 = <em>$ 42 if this tank was full provided a community cost of $ 2.8 per gallon</em>

5. Now this tank is 20% full, so we must calculate the cost to fill the other 80% up. That would be 80/100 * 42 = 4/5 * 42 = 168/5 = <em>$ 33.6 to fill this tank, provided a community cost of $2.8 per gallon</em>

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Let theta be an angle in quadrant II such that cos theta = -2/3
Iteru [2.4K]

Answer:

So we have \csc(\theta)=\frac{3 \sqrt{5}}{5} \text{ and } \tan(\theta)=\frac{-\sqrt{5}}{2}.

Step-by-step explanation:

Ok so we are in quadrant 2, that means sine is positive while cosine is negative.

We are given \cos(\theta)=\frac{-2}{3}(\frac{\text{adjacent}}{\text{hypotenuse}}).

So to find the opposite we will just use the Pythagorean Theorem.

a^2+b^2=c^2

(2)^2+b^2=(3)^2

4+b^2=9

b^2=5

b=\sqrt{5}  This is the opposite side.

Now to find \csc(\theta) and \tan(\theta).

\csc(\theta)=\frac{\text{hypotenuse}}{\text{opposite}}=\frac{3}{\sqrt{5}}.

Some teachers do not like the radical on bottom so we will rationalize the denominator by multiplying the numerator and denominator by sqrt(5).

So \csc(\theta)=\frac{3}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}=\frac{3 \sqrt{5}}{5}.

And now \tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}=\frac{\sqrt{5}}{-2}=\frac{-\sqrt{5}}{2}.

So we have \csc(\theta)=\frac{3 \sqrt{5}}{5} \text{ and } \tan(\theta)=\frac{-\sqrt{5}}{2}.

7 0
3 years ago
Read 2 more answers
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Sophie [7]
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3 years ago
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Answer:

6.75

Step-by-step explanation:

use the proportion:

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8 0
1 year ago
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4 years ago
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