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marishachu [46]
2 years ago
11

10: A stadium brings in $16.25 million per year. It pays football-related expenses of $13.5 million and stadium expenses of $2.7

million per year. What is the stadium's current profit margin?
Mathematics
1 answer:
Oksanka [162]2 years ago
5 0

The stadium brings is $16,250,000 per year. (Revenue)

Football-related expenses equal $13,500,000 per year. (Cost)

Stadium expenses equal $2,700,000 per year. (Cost)

Profit margin is calculated by dividing net profit with the revenue.

To get net profit, subtract the revenue with the total cost.

First, add the two costs together to get the total cost.

$13,500,000 + $2,700,000 = $16,200,000 (total cost)

$16,250,000 (revenue) - $16,200,000 (total cost) = $50,000 (net profit)

$50,000 (net profit) ÷ $16,250,000 (revenue) = 0.0030769

To turn that decimal into a fraction, multiply it with 100

0.0030769 x 100 = 0.30769%

Round this up to 0.308%

The stadium's current profit margin is 0.308%

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Answer:

Step-by-step explanation:

Lines m and l are the parallel lines and a line 'n' is a transverse intersecting these lines.

m∠2 = 50°

m∠1 + m∠2 = 180° [Linear pair of angles]

m∠1 = 180° - 50°

m∠1 = 130°

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m∠3 + m∠5 = 180° [Consecutive interior angles]

m∠5 = 180° - m∠3

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        = 50°

m∠6 + m∠5 = 180° [Linear pair of angles]

m∠6 = 180° - 50° = 130°

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A person is standing 50 ft from a statue. The person looks up at an angle of elevation of 16o when staring at the top of the sta
NemiM [27]

Answer:

The height of the statue is 21.4 feet

Step-by-step explanation:

We are given

A person is standing 50 ft from a statue

The person looks up at an angle of elevation of 16 degree when staring at the top of the statue

hen the person looks down at an angle of depression of 8 degree when staring at the base of the statue

Firstly, we will draw diagram

In triangle ABF:

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we can use trig formula

tan(16)=\frac{y}{50}

y=50tan(16)

y=14.33727ft

now, we can find x

In triangle DEF:

we can use trig formula

tan(8)=\frac{x}{50}

x=50tan(8)

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Match the identities to their values taking these conditions into consideration sinx=sqrt2 /2 cosy=-1/2 angle x is in the first
BaLLatris [955]

Answer:

\cos(x+y) goes with -\frac{\sqrt{6}+\sqrt{2}}{4}

\sin(x+y) goes with \frac{\sqrt{6}-\sqrt{2}}{4}

\tan(x+y) goes with \sqrt{3}-2

Step-by-step explanation:

\cos(x+y)

\cos(x)\cos(y)-\sin(x)\sin(y) by the addition identity for cosine.

We are given:

\sin(x)=\frac{\sqrt{2}}{2} which if we look at the unit circle we should see

\cos(x)=\frac{\sqrt{2}}{2}.

We are also given:

\cos(y)=\frac{-1}{2} which if we look the unit circle we should see

\sin(y)=\frac{\sqrt{3}}{2}.

Apply both of these given to:

\cos(x+y)

\cos(x)\cos(y)-\sin(x)\sin(y) by the addition identity for cosine.

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\frac{-\sqrt{2}-\sqrt{6}}{4}

-\frac{\sqrt{6}+\sqrt{2}}{4}

Apply both of the givens to:

\sin(x+y)

\sin(x)\cos(y)+\sin(y)\cos(x) by addition identity for sine.

\frac{\sqrt{2}}{2}\frac{-1}{2}+\frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2}

\frac{-\sqrt{2}+\sqrt{6}}{4}

\frac{\sqrt{6}-\sqrt{2}}{4}

Now I'm going to apply what 2 things we got previously to:

\tan(x+y)

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-(2-\sqrt{3})

Distribute:

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6 0
1 year ago
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