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irina [24]
3 years ago
10

A tank has 14 gallons of water in it when all of a sudden the water begins draining from the tank. Recall that water weights 8.3

45 pounds per gallon.
If 12 pounds of water have left the tank how many gallons of water are remaining
Mathematics
1 answer:
andreev551 [17]3 years ago
7 0
14 is the initial amount of water.

12 pounds is what has left.

Use a ratio
8.345 pounds per 1 gallon equals 12 pounds per x gallons.
8.345/1=12/x
x=12/8.345
x=1.438 gallons

Subtract this number from the initial to get the remaining amount.
14-1.438=12.562

12.562 gallons of water are remaining.
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NISA [10]

Answer:

\theta = 108.29

Step-by-step explanation:

Given

u =

v =

Required:

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The angle \theta is calculated as thus:

cos\theta = \frac{u.v}{|u|.|v|}

For a vector

A =

A = a * b

cos\theta = \frac{u.v}{|u|.|v|} becomes

cos\theta = \frac{.}{|u|.|v|}

cos\theta = \frac{2*6+4*-7}{|u|.|v|}

cos\theta = \frac{12-28}{|u|.|v|}

cos\theta = \frac{-16}{|u|.|v|}

For a vector

A =

|A| = \sqrt{a^2 + b^2}

So;

|u| = \sqrt{2^2 + 6^2}

|u| = \sqrt{4 + 36}

|u| = \sqrt{40}

|v| = \sqrt{4^2+(-7)^2}

|v| = \sqrt{16+49}

|v| = \sqrt{65}

So:

cos\theta = \frac{-16}{|u|.|v|}

cos\theta = \frac{-16}{\sqrt{40}*\sqrt{65}}

cos\theta = \frac{-16}{\sqrt{2600}}

cos\theta = \frac{-16}{\sqrt{100*26}}

cos\theta = \frac{-16}{10\sqrt{26}}

cos\theta = \frac{-8}{5\sqrt{26}}

Take arccos of both sides

\theta = cos^{-1}(\frac{-8}{5\sqrt{26}})

\theta = cos^{-1}(\frac{-8}{5 * 5.0990})

\theta = cos^{-1}(\frac{-8}{25.495})

\theta = cos^{-1}(-0.31378701706)

\theta = 108.288386087

<em></em>\theta = 108.29<em> (approximated)</em>

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