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pantera1 [17]
3 years ago
7

if employee a logged 120 hours of travel in four years and employee b logged of travel at a 20% higher rate how many more hours

did employee be have that an employee after 1.5 years
Mathematics
1 answer:
Gnoma [55]3 years ago
6 0
First we need to find the rate of logging for employee A per year. To do this, the number of hours can be divided by the number of years to find out how many hours of travel they log per year.

120 ÷ 4 = 30. Employee A logs roughly 30 hours of travel per year.

To find a percentage increase of 20%, you need to multiply by 1.2. 
30 x 1.2 = 36. 
Therefore, Employee B logs roughly 36 hours of travel per year.

To find out how many hours of travel Employee B logs after 1.5 years, we simply need to multiply 36 by 1.5, which equals 54.

To find out how many hours of travel Employee A logs after 1.5 years, we simply need to multiply 30 by 1.5, which equals 45.

Finally, we need to find out how many MORE hours Employee B logged after 1.5 years compared to Employee A. To do this, we simply need to do 54 - 45 = 9 hours.

Working = (((120/4) x 1.2) x 1.5) - (120/4 x 1.5)
Answer = 9 hours.
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we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now

\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\\
b=opposite\\
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\\\\\\
\pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a\implies \stackrel{IV~quadrant}{+\sqrt{35}=a}

recall that 

\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
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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,

\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}
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