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3 years ago

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Byron can fill a drying rack with dishes in 15 minutes. While Byron fills the drying rack, Emmitt takes dishes out of the drying

rack and puts them in bins. With Emmitt taking dishes out of the drying rack, it takes 35 minutes for Byron to fill the drying rack.
Which of the following can be used to determine the amount of time it takes for Emmitt to empty the drying rack if Byron is not adding dishes?

Choices -

1 over 15 minus 1 over 35 equals 1 over x
1 over 15 minus 1 over x equals 1 over 35
1 over 35 minus 1 over x equals 1 over 15
1 over 15 minus 1 over x equals x over 35

1 answer:

Olegator [25]3 years ago

7 0

Answer:

Hi!

The correct answer is d) 1 over 15 minus 1 over x equals x over 35.

Step-by-step explanation:

We know the amount of time that Byron can fill a drying rack with dishes: 15 minutes.

We know the amount of time that Byron can fill a drying rack with dishes while Emmitt is taking dishes out of the drying rack: 35 minutes.

We don't know the amount of time that Emmitt can take dishes out of the drying rack: x minutes.

So, if Byron is filling the drying rack in 15 minutes and Emmitt is taking out the dishes at x minutes, it takes 35 minutes to fill the drying rack.

$\frac{1}{15}-\frac{1}{x} = \frac{1}{35}$ <em>// multiply by 105x both sides.</em>

$105x\frac{1}{15}-105x\frac{1}{x} =105x \frac{1}{35}$ <em>// simplifly</em>

$7x - 105 = 3x$ <em> //substract 3x and add 105 in both sides</em>

$7x - 3x - 105 + 105 = 3x - 3x + 105$ <em>// solve</em>

$4x = 105$ <em>//</em> <em>divide by 4 in both sides</em>

$\frac{4x}{4} = \frac{105}{4}$ <em>// solve</em>

$x = 26.25$

It makes sense that Byron fills the rack faster than Emmitt empties the rack.

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$\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
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\end{cases}
\\\\\\
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$\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad\qquad 
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\\\\\\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\\\\\\
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
\qquad \qquad 
% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}$

therefore, let's just plug that on the remaining ones,

$\bf sin(\theta)=\cfrac{-1}{6}
\qquad\qquad 
cos(\theta)=\cfrac{\sqrt{35}}{6}
\\\\\\
% tangent
tan(\theta)=\cfrac{-1}{\sqrt{35}}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{\sqrt{35}}{1}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}$

now, let's rationalize the denominator on tangent and secant,

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\\\\\\
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