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vaieri [72.5K]
3 years ago
5

Sheila loves to eat chicken tenders. A small order comes with 5 chicken

Mathematics
1 answer:
aev [14]3 years ago
5 0

Answer:  The required number of large order of chicken tenders is 5.

Step-by-step explanation:  Given that Sheila loves to eat chicken tenders. A small order comes with 5 chicken  tenders, and a large order comes with 8 chicken tenders.

Last month, Sheila ordered chicken tenders a total of 7 times. She received a total of 50 chicken tenders.

We are to find the number of large chicken tenders received by Sheila.

Let x and y represents the number of small orders and large orders respectively of chicken tenders.

Then, according to the given information, we have

x+y=7\\\\\Rightarrow x=7-y~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

and

5x+8y=50\\\\\Rightarrow 5(7-y)+8y=50~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Using equation (i)}]\\\\\Rightarrow 35-5y+8y=50\\\\\Rightarrow 3y=50-35\\\\\Rightarrow 3y=15\\\\\Rightarrow y=\dfrac{15}{3}\\\\\Rightarrow y=5.

Thus, the required number of large order of chicken tenders is 5.

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Answer:

A = 74.7^\circ

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C = 62.8^\circ

Step-by-step explanation:

Given

A = (-1,2) \to (x_1,y_1)

B = (2,8) \to (x_2,y_2)

C = (4,1) \to (x_3,y_3)

Required

The measure of each angle

First, we calculate the length of the three sides of the triangle.

This is calculated using distance formula

d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2

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d = \sqrt{(-1 - 2)^2 + (2 - 8)^2

d = \sqrt{(-3)^2 + (-6)^2

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So:

AB = \sqrt{45

For BC

B = (2,8) \to (x_2,y_2)

C = (4,1) \to (x_3,y_3)

BC = \sqrt{(2 - 4)^2 + (8 - 1)^2

BC = \sqrt{(-2)^2 + (7)^2

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A = (-1,2) \to (x_1,y_1)

C = (4,1) \to (x_3,y_3)

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So, we have:

AB = \sqrt{45

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AC = \sqrt{26

By representation

AB \to c

BC \to a

AC \to b

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a = \sqrt{53

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By cosine laws, the angles are calculated using:

a^2 = b^2 + c^2 -2bc \cos A

b^2 = a^2 + c^2 -2ac \cos B

c^2 = a^2 + b^2 -2ab\ cos C

a^2 = b^2 + c^2 -2bc \cos A

(\sqrt{53})^2 = (\sqrt{26})^2 +(\sqrt{45})^2 - 2 * (\sqrt{26}) +(\sqrt{45}) * \cos A

53 = 26 +45 - 2 * 34.21 * \cos A

53 = 26 +45 - 68.42 * \cos A

Collect like terms

53 - 26 -45 = - 68.42 * \cos A

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Solve for \cos A

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Take arc cos of both sides

A =\cos^{-1}(0.2631)

A = 74.7^\circ

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26 = 53 +45 -97.67 * \cos B

Collect like terms

26 - 53 -45= -97.67 * \cos B

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Solve for \cos B

\cos B = \frac{-72}{-97.67}

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Take arc cos of both sides

B = \cos^{-1}(0.7372)

B = 42.5^\circ

For the third angle, we use:

A + B + C = 180 --- angles in a triangle

Make C the subject

C = 180 - A -B

C = 180 - 74.7 -42.5

C = 62.8^\circ

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Note:
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