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Norma-Jean [14]

2 years ago

15

There was a fish tank for sale at the City Auction. At the beginning of the auction it had 10 gallons of water in it, Thirty min

utes later, 1 cup of water had leaked out . How much water was left in the tank?

1 answer:

Komok [63]2 years ago

7 0

Answer: 9 gallons, 15 cups

Step-by-step explanation:

Note that 1 gallon = 16 cups

Since at the beginning of the auction, the tank had 10 gallons of water in it, and thirty minutes later, 1 cup of water had leaked out .

The amount of water that was left in the tank will be:

= 10 gallons - 1 cup

= 9 gallons, 15 cups

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Given two points P(sinθ+2, tanθ-2) and Q(4sin²θ+4sinθcosθ+2acosθ, 3sinθ-2cosθ+a). Find constant "a" and the corresponding value

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Answer:

$\rm\displaystyle \displaystyle \displaystyle θ= {60}^{ \circ} , {300}^{ \circ}$

$\rm \displaystyle a = - \frac{ \sqrt{3} }{2} - 1, \frac{\sqrt{3}}{2} - 1$

Step-by-step explanation:

we are given two <u>coincident</u><u> points</u>

$\displaystyle P( \sin(θ)+2, \tan(θ)-2) \: \text{and } \\ \displaystyle Q(4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)$

since they are coincident points

$\rm \displaystyle P( \sin(θ)+2, \tan(θ)-2) = \displaystyle Q(4 \sin ^{2} (θ)+4 \sin(θ )\cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)$

By order pair we obtain:

$\begin{cases} \rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ) = \sin( \theta) + 2 \\ \\ \displaystyle 3 \sin( \theta) - 2 \cos( \theta) + a = \tan( \theta) - 2\end{cases}$

now we end up with a simultaneous equation as we have two variables

to figure out the simultaneous equation we can consider using <u>substitution</u><u> method</u>

to do so, make a the subject of the equation.therefore from the second equation we acquire:

$\begin{cases} \rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sinθ \cos(θ)+2a \cos(θ )= \sin( \theta) + 2 \\ \\ \boxed{\displaystyle a = \tan( \theta) - 2 - 3 \sin( \theta) + 2 \cos( \theta) } \end{cases}$

now substitute:

$\rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2 \cos(θ) \{\tan( \theta) - 2 - 3 \sin( \theta) + 2 \cos( \theta) \}= \sin( \theta) + 2$

distribute:

$\rm\displaystyle \displaystyle 4 \sin ^{2}( θ)+4 \sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) - 6 \sin( \theta) \cos( \theta) + 4 \cos ^{2} ( \theta) = \sin( \theta) + 2$

collect like terms:

$\rm\displaystyle \displaystyle 4 \sin ^{2}( θ) - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) + 4 \cos ^{2} ( \theta) = \sin( \theta) + 2$

rearrange:

$\rm\displaystyle \displaystyle 4 \sin ^{2}( θ) + 4 \cos ^{2} ( \theta) - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) + = \sin( \theta) + 2$

by <em>Pythagorean</em><em> theorem</em> we obtain:

$\rm\displaystyle \displaystyle 4 - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) = \sin( \theta) + 2$

cancel 4 from both sides:

$\rm\displaystyle \displaystyle - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) = \sin( \theta) - 2$

move right hand side expression to left hand side and change its sign:

$\rm\displaystyle \displaystyle - 2\sin(θ) \cos(θ)+\sin(θ ) - 4\cos( \theta) + 2 = 0$

factor out sin:

$\rm\displaystyle \displaystyle \sin (θ) (- 2 \cos(θ)+1) - 4\cos( \theta) + 2 = 0$

factor out 2:

$\rm\displaystyle \displaystyle \sin (θ) (- 2 \cos(θ)+1) + 2(- 2\cos( \theta) + 1 ) = 0$

group:

$\rm\displaystyle \displaystyle ( \sin (θ) + 2)(- 2 \cos(θ)+1) = 0$

factor out -1:

$\rm\displaystyle \displaystyle - ( \sin (θ) + 2)(2 \cos(θ) - 1) = 0$

divide both sides by -1:

$\rm\displaystyle \displaystyle ( \sin (θ) + 2)(2 \cos(θ) - 1) = 0$

by <em>Zero</em><em> product</em><em> </em><em>property</em> we acquire:

$\begin{cases}\rm\displaystyle \displaystyle \sin (θ) + 2 = 0 \\ \displaystyle2 \cos(θ) - 1= 0 \end{cases}$

cancel 2 from the first equation and add 1 to the second equation since -1≤sinθ≤1 the first equation is false for any value of theta

$\begin{cases}\rm\displaystyle \displaystyle \sin (θ) \neq - 2 \\ \displaystyle2 \cos(θ) = 1\end{cases}$

divide both sides by 2:

$\rm\displaystyle \displaystyle \displaystyle \cos(θ) = \frac{1}{2}$

by unit circle we get:

$\rm\displaystyle \displaystyle \displaystyle θ= {60}^{ \circ} , {300}^{ \circ}$

so when θ is 60° a is:

$\rm \displaystyle a = \tan( {60}^{ \circ} ) - 2 - 3 \sin( {60}^{ \circ} ) + 2 \cos( {60}^{ \circ} )$

recall unit circle:

$\rm \displaystyle a = \sqrt{3} - 2 - \frac{ 3\sqrt{3} }{2} + 2 \cdot \frac{1}{2}$

simplify which yields:

$\rm \displaystyle a = - \frac{ \sqrt{3} }{2} - 1$

when θ is 300°

$\rm \displaystyle a = \tan( {300}^{ \circ} ) - 2 - 3 \sin( {300}^{ \circ} ) + 2 \cos( {300}^{ \circ} )$

remember unit circle:

$\rm \displaystyle a = - \sqrt{3} - 2 + \frac{3\sqrt{ 3} }{2} + 2 \cdot \frac{1}{2}$

simplify which yields:

$\rm \displaystyle a = \frac{ \sqrt{3} }{2} - 1$

and we are done!

disclaimer: also refer the attachment I did it first before answering the question

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