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2 years ago

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A clothing manufacturer produced 900 items with flaws. Each flawed item represented a loss of $12.50. Which expression represent

s one way to calculate the manufacturer’s total loss?
A. 900 x (-$12.50)
B. 900 + (-$12.50)
C. 900 - (-$12.50)
D. 900 ÷ (-$12.50)

1 answer:

OverLord2011 [107]2 years ago

7 0

A
900 a piece
negative 12.50 because it's a loss of money

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Wilton collected $8.50 in dimes and quarters while cleaning his father's truck. If the number of quarters is 3 times the number

dusya [7]

Answer:

Wilton found 10 dimes while cleaning his father's truck.

Step-by-step explanation:

Given that:

Total amount collected = $8.50 = 8.50*100 = 850 cents

Let,

x be the number of dimes

y be the number of quarters

According to given statement,

10x+25y=850 Eqn 1

y = 3x Eqn 2

Putting y=3x in Eqn 1

$10x+25(3x)=850\\10x+75x=850\\85x=850$

Dividing both sides by 85

$\frac{85x}{85}=\frac{850}{85}\\x=10$

As x represents dimes, therefore, number of dimes is 10.

Hence,

Wilton found 10 dimes while cleaning his father's truck.

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2 years ago

Find the remaining trigonometric ratios of θ if csc(θ) = -6 and cos(θ) is positive

VikaD [51]

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}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
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recall that

$\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad\qquad 
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\\\\\\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\\\\\\
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
\qquad \qquad 
% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}$

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

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

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

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

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Step-by-step explanation:

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