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Xelga [282]
3 years ago
9

The football team is selling hats to raise $425 for a trip they have $175 already if the team member sell the hats for $12 each

at least how many hats do they need to sell to make enough money for their trip.
Mathematics
1 answer:
xz_007 [3.2K]3 years ago
4 0

Answer:

The minimum number of hats which team need to sell t make enough money for the trip is 21 .

Step-by-step explanation:

Given as :

The money which team needs for the trip = $ 425

The money which team already = $ 175

So, The rest money which team need = $ 425 - $ 175 = $ 250

Now, The sell price for the hats = $ 12

Let The minimum number of hats which team need to sell = x

So, According to question

∵ The rest money which team need = $ 250

The selling price of each hats = $ 12

Or,  The minimum number of hats which team need to sell = \dfrac{\textrm Rest money which team needs}{\textrm The selling price of each hats}

Or, x = \frac{250}{12}

∴  x = \frac{250}{12}

I.e x = 20.8

So, The minimum number of hats need to be sell = 20.8 ≈ 21

Hence The minimum number of hats which team need to sell t make enough money for the trip is 21 . Answer

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<img src="https://tex.z-dn.net/?f=cos%20%5B%20arctan%28%5Cfrac%7B12%7D%7B5%7D%29%20-%20arcsin%20%28%5Cfrac%7B-3%7D%7B5%7D%29%5D"
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Answer: -16/65

Step-by-step explanation:

Drawing the right triangle (as attached) gives us that \arctan \left(\frac{12}{5} \right)=\arcsin \left(\frac{12}{13} \right)

Also, -\arcsin \left(-\frac{3}{5} \right)=\arcsin \left(\frac{3}{5} \right)

This means our original expression is equal to:

\cos \left[\arcsin \left(\frac{12}{13} \right)+\arcsin \left(\frac{3}{5} \right) \right]

Using the cosine addition formula, which states \cos(a+b)=\cos a \cos b-\sin a \sin b, we get this itself is equal to:

\cos \left(\arcsin \left(\frac{12}{13} \right) \right)\cos \left(\arcsin \left(\frac{3}{5} \right)\right)-\sin \left(\arcsin \left(\frac{12}{13} \right) \right)\sin \left(\arcsin \left(\frac{3}{5} \right)\right)

Since \sin^{2} \theta+\cos^{2} \theta=1, we know that:

\sin^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)+\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=1\\\\\frac{144}{169} +\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=1\\\\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=\frac{25}{169}\\\\cos \left(\arcsin \left(\frac{12}{13} \right)\right)=\frac{5}{13}

Similarly, cos(arcsin(3/5))=4/5.

This means the given expression is equal to:

\left(\frac{5}{13} \right) \left(\frac{4}{5} \right)-\left(\frac{12}{13} \right) \left(\frac{3}{5} \right)\\\\\frac{20}{65}-\frac{36}{65}=\boxed{-\frac{16}{65}}

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Step-by-step explanation:

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Answer:

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