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V125BC [204]
3 years ago
9

someone came into my shop and stole $100 out of the register. the same person came back 5 min later with the same $100 and purch

ased $70 dollars worth of items and i gave them back $30 dollars in change. how much money did i lose?
Mathematics
2 answers:
marishachu [46]3 years ago
8 0
You lost $100 at the beginning so your balance is -100
Now he came in and bought $70 worth of stuff which is -100 + 70 = -30 
but you gave him -30 back so -30 - 30 = -60

You lost 60 dollars.
Ivahew [28]3 years ago
7 0
$30, because you were robbed of $100, then given back $100, then ou gave back $30 after keeping $70 of the $100. Hope that made sense. But basically, -100, +100, -30.
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Find the remaining trigonometric ratios of θ if csc(θ) = -6 and cos(θ) is positive
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Now, the cosecant of θ is -6, or namely -6/1.

however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.

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}\\\\
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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 
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\\\\\\
% tangent
tan(\theta)=\cfrac{-1}{\sqrt{35}}
\qquad \qquad 
% cotangent
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\\\\\\
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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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