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zepelin [54]
3 years ago
13

A factory operates 5 days in a week. In particular week 5% of the total number of glasses produced, 419 glasses, are found crack

ed. What is the average number of glasses produced per day in that week?​
Mathematics
1 answer:
Novosadov [1.4K]3 years ago
4 0

Answer:

the answer would be 1676 glasses per day

Step-by-step explanation:

if you set up and equation to have the total number of glasses that week, you'd have 5/100 = 419/x

you cross multiply and solve to get x=8380. that is the total number of glasses produced in a week.

since you want the number of glasses produced per day, you'd have to divide that by the number of operating days in a week, which in this case is 5.

after dividing 8380 by 5, you would have an end result of 1676.

Hope this helps!

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Can someone please explain how to do this!!!
DedPeter [7]

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!

8 0
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