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rjkz [21]
3 years ago
11

You have a small business making and delivering box lunches. You calculate your average weekly cost y of producing x lunches usi

ng the function y = 2.1x + 75. Last week your cost was $600. How many lunches did you make last week?
Mathematics
1 answer:
kogti [31]3 years ago
6 0

250 lunches are produced by the small business in last week.

<u>Step-by-step explanation:</u>

It is given that,

  • y ⇒ the average cost per week.
  • x ⇒ the number of lunches produced per week.

The function relating these two factors x and y is given as y = 2.1x + 75

  • The cost of the last week is y = $600.
  • The lunches made last week is x = unknown.

<u>To find the value of x :</u>

Substitute y= 600 in the given function,

⇒ 600 = 2.1x + 75

⇒ 2.1x = 600 - 75

⇒ x = 525 / 2.1

⇒ x = 250

Therefore, the lunches prepared last week is 250.

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Hi there!

So we are given that:-

  • tan theta = 7/24 and is on the third Quadrant.

In the third Quadrant or Quadrant III, sine and cosine both are negative, which makes tangent positive.

Since we want to find the value of cos theta. cos must be less than 0 or in negative.

To find cos theta, we can either use the trigonometric identity or Pythagorean Theorem. Here, I will demonstrate two ways to find cos.

<u>U</u><u>s</u><u>i</u><u>n</u><u>g</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>I</u><u>d</u><u>e</u><u>n</u><u>t</u><u>i</u><u>t</u><u>y</u>

\large \displaystyle{ {tan}^{2}  \theta + 1 =  {sec}^{2}  \theta}

Substitute tan theta = 7/24 in.

\large \displaystyle{ {(  \frac{7}{24}) }^{2}  + 1 =  {sec}^{2} \theta }

Evaluate.

\large \displaystyle{ \frac{49}{576}  + 1 =  {sec}^{2}  \theta} \\  \large \displaystyle{ \frac{625}{576}  =  {sec}^{2}  \theta}

Reminder -:

\large \displaystyle{ sec \theta =  \frac{1}{cos \theta} }

Hence,

\large \displaystyle{ \frac{576}{625}  =  {cos}^{2}  \theta} \\  \large \displaystyle{ \sqrt{ \frac{576}{625} }  = cos  \theta} \\  \large \displaystyle{ \frac{24}{25}  = cos \theta}

Because in QIII, cos<0. Hence,

\large \displaystyle \boxed{ \blue{cos \theta =  -  \frac{24}{25} }}

<u>U</u><u>s</u><u>i</u><u>n</u><u>g</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>P</u><u>y</u><u>t</u><u>h</u><u>a</u><u>g</u><u>o</u><u>r</u><u>e</u><u>a</u><u>n</u><u> </u><u>T</u><u>h</u><u>e</u><u>o</u><u>r</u><u>e</u><u>m</u>

\large \displaystyle{ {a}^{2}  +  {b}^{2}  =  {c}^{2} }

Define c as our hypotenuse while a or b can be adjacent or opposite.

Because tan theta = opposite/adjacent. Therefore:-

\large \displaystyle{ {7}^{2}  +  {24}^{2}  =  {c}^{2} } \\  \large \displaystyle{49 + 576 =  {c}^{2} } \\  \large \displaystyle{625 =  {c}^{2} } \\  \large \displaystyle{25 = c}

Thus, the hypotenuse side is 25. Using the cosine ratio:-

\large \displaystyle{cos \theta =  \frac{adjacent}{hypotenuse}}

Therefore:-

\large \displaystyle{cos \theta =  \frac{24}{25} }

Because cos<0 in Q3.

\large \displaystyle \boxed{ \red{cos \theta =  -  \frac{24}{25} }}

Hence, the value of cos theta is -24/25.

Let me know if you have any questions!

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