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katen-ka-za [31]
3 years ago
14

A clothing store is running a special in which customers receive a free pair of jeans with the purchase of 4 shirts. It costs th

e clothing company $3.46 to produce one shirt and $5.56 to produce one pair of jeans. The store is selling each shirt for $21.46. Approximately how much money will the company make if 42 customers take advantage of the special?
Mathematics
1 answer:
ahrayia [7]3 years ago
6 0
If one shirt equals to $21.46, and you need to purchase 4 shirts to get the free pair of jeans, you multiply 4 by 21.46 and get 85.84. But if there is 42 customers taking advantage of the special, then you need to multiply 85.84 by 42, and you get 3,605.28.
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Slav-nsk [51]
The pricing of the boxes seems to be consistently 2.40 per box. Ethan's cost is
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5 0
3 years ago
Find the area of the region that lies inside the first curve and outside the second curve.
marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the  the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = \dfrac{1}{2}

\theta = -\dfrac{\pi}{3}, \ \  \dfrac{\pi}{3}

Now, the area of the  region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \  \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \  5^2 d \theta

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \  \theta  d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \   d \theta

A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  \dfrac{cos \ 2 \theta +1}{2}  \end {pmatrix} \ \ d \theta - \dfrac{25}{2}  \begin {bmatrix} \theta   \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}

A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  {cos \ 2 \theta +1}  \end {pmatrix} \ \    d \theta - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}}    \ \ - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{2 \pi}{3} \end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3})  \end {bmatrix} - \dfrac{25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} +   \dfrac{\pi}{3}  \end {bmatrix}- \dfrac{ 25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3}   \end {bmatrix}- \dfrac{ 25 \pi}{3}

A =    \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx
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