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leva [86]
4 years ago
11

Pine Bluff Middle School is having its annual Spring Fling dance, which will cost

Mathematics
1 answer:
Anon25 [30]4 years ago
3 0
I think your asking  how much the school will have left over. If so:

125 ( not including the tickets)
It depends on how many kids are coming to the dance in order to fully solve this problem.

You didn't word this properly so that what I got with what I had:)

If you have anymore questions, ask me them on my profile so I'll be sure to get them:)

I hope this helps:)
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Prove that: (secA-cosec A) (1+cot A +tan A) =( sec^2A/cosecA)-(Cosec^2A/secA)<br>​
Ksju [112]

Step-by-step explanation:

(\sec A - \csc A)(1 + \cot A + \tan A)

=(\sec A - \csc A)\left(1 + \dfrac{\cos A}{\sin A} + \dfrac{\sin A}{\cos A} \right)

=(\sec A - \csc A)\left(1 + \dfrac{\cos^2 A + \sin^2 A}{\sin A\cos A} \right)

=(\sec A - \csc A)\left(\dfrac{1 + \sin A \cos A}{\sin A \cos A} \right)

=\left(\dfrac{\frac{1}{\cos A} - \frac{1}{\sin A}+\sin A - \cos A}{\sin A\cos A}\right)

=\dfrac{\sin A - \sin A \cos^2A - \cos A + \cos A\sin^2A}{(\sin A\cos A)^2}

=\dfrac{\sin A(1 - \cos^2A) - \cos A (1 - \sin^2 A)}{(\sin A\cos A)^2}

=\dfrac{\sin^3A - \cos^3A}{\sin^2A\cos^2A}

=\dfrac{\sin A}{\cos^2A} - \dfrac{\cos A}{\sin^2A}

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5 0
3 years ago
the scale from a square park to a drawing of the park is 5 m to 1 cm the actual park has an area of 1,600 m² what is the area of
ludmilkaskok [199]

Answer:

<h2>64 cm²</h2>

Step-by-step explanation:

given:

the scale from a square park to a drawing of the park is 5 m to 1 cm

the actual park has an area of 1,600 m²

find:

what is the area of the drawing of the park​

solution:

drawing scale is 5m : 1cm

<u>1,600 m² </u> = <u>      5² m²  </u>    

   x                  1² cm²

cross multiply:

5² m² x = 1600 m² (1 cm²)

x =<u> 1600 m² </u>

      5² m²

x = 64 cm²

therefore, the area of a square park in the drawing is 64 cm²

5 0
3 years ago
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