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almond37 [142]
3 years ago
10

Suppose you invest $10,400 in equipment to manufacture a new board game. Each game costs $2.65 to manufacture and sells for $20.

How many games must you sell to break even?
Mathematics
1 answer:
tamaranim1 [39]3 years ago
3 0
Answer: 599.42 games
To answer this question, you need to calculate the revenue for every game sold first. The calculation would be:
Revenue= sell cost - manufacture cost= $20-$2.65= $17.35/game.

Break even is the point when the game manufacturing cost(equipment cost included) is covered by the revenue you got. Then the calculation would be:

total game cost= revenue per game * number of game sold
$10400= number of game sold * $17.35
number of game sold= 10400/17.35=  599.42 games

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Does the order pair, (2,3) satisfy the following function y=-3x-2 Yes,No
grin007 [14]

Answer:

No

Step-by-step explanation:

The function y=-3x-2 does not go through the ordered pair (2,3).

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N the spring of 2008, petrol cost £1.029 per litre in london. on the same day, the exchange rate was $1 = £0.497. what was the p
Dmitriy789 [7]

1 liter of petrol in London costs £1.029

Since 1 gallon consist of 3.7854 liters, the cost of 3.7854 liters is calculated as:

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Next, we convert the pounds (£) to dollar($)

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3 0
3 years ago
Use the given information to find the exact value of the trigonometric function
eimsori [14]
\begin{gathered} \csc \theta=-\frac{6}{5} \\ \tan \theta>0 \\ \cos \frac{\theta}{2}=\text{?} \end{gathered}

Half Angle Formula

\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\  \\  \end{gathered}

Answer:

\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}

Checking:

\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{  (3rd quadrant)} \end{gathered}

Also,

\csc \theta=\frac{1}{\sin\theta}=\frac{1}{\sin (236.44)}=-\frac{6}{5}\text{ QED}

The answer is none of the choices

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Vinil7 [7]
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