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d1i1m1o1n [39]
3 years ago
8

) 2 ft 2 ft Perimeter: Area:

Mathematics
1 answer:
vitfil [10]3 years ago
3 0

Answer:

<h3>perimeter = 8 ft</h3><h3>area = 4 ft</h3>

Solution:

perimeter- 2 + 2 + 2 + 2 = 8

area of square- (length × width) 2 × 2 = 4

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What are the exact values of sin(2π3radians) and cos(2π3radians)
deff fn [24]
We know that
2π/3 radians-------> convert to degrees-----> 2*180/3---> 120°
120°=90°+30°

Part a) Find <span>sin(2π/3)
</span>sin(2π/3)=sin (90°+30°)
we know that
sin (A+B)=sin A*cos B+cos A*sin B
so
sin (90°+30°)=sin 90*cos 30+cos 90*sin 30
sin 90=1
cos 30=√3/2
cos 90=0
sin 30=1/2
sin (90°+30°)=1*√3/2+0*1/2-----> √3/2

the answer part a) is
sin(2π/3)=√3/2

Part b) Find cos (2π/3)
cos (2π/3)=cos (90°+30°)
we know that
cos (A+B)=cos A*cos B-sin A*sin B
so
cos (90°+30°)=cos 90*cos 30-sin 90*sin 30
sin 90=1
cos 30=√3/2
cos 90=0
sin 30=1/2
cos (90°+30°)=0*√3/2-1*1/2----> -1/2

the answer part b) is
cos (2π/3)=-1/2
8 0
3 years ago
A company manufactures and sells x television sets per month. The monthly cost and​ price-demand equations are ​C(x)equals72 com
solmaris [256]

Answer:

Part (A)

  • 1. Maximum revenue: $450,000

Part (B)

  • 2. Maximum protit: $192,500
  • 3. Production level: 2,300 television sets
  • 4. Price: $185 per television set

Part (C)

  • 5. Number of sets: 2,260 television sets.
  • 6. Maximum profit: $183,800
  • 7. Price: $187 per television set.

Explanation:

<u>0. Write the monthly cost and​ price-demand equations correctly:</u>

Cost:

      C(x)=72,000+70x

Price-demand:

     

      p(x)=300-\dfrac{x}{20}

Domain:

        0\leq x\leq 6000

<em>1. Part (A) Find the maximum revenue</em>

Revenue = price × quantity

Revenue = R(x)

           R(x)=\bigg(300-\dfrac{x}{20}\bigg)\cdot x

Simplify

      R(x)=300x-\dfrac{x^2}{20}

A local maximum (or minimum) is reached when the first derivative, R'(x), equals 0.

         R'(x)=300-\dfrac{x}{10}

Solve for R'(x)=0

      300-\dfrac{x}{10}=0

       3000-x=0\\\\x=3000

Is this a maximum or a minimum? Since the coefficient of the quadratic term of R(x) is negative, it is a parabola that opens downward, meaning that its vertex is a maximum.

Hence, the maximum revenue is obtained when the production level is 3,000 units.

And it is calculated by subsituting x = 3,000 in the equation for R(x):

  • R(3,000) = 300(3,000) - (3000)² / 20 = $450,000

Hence, the maximum revenue is $450,000

<em>2. Part ​(B) Find the maximum​ profit, the production level that will realize the maximum​ profit, and the price the company should charge for each television set. </em>

i) Profit(x) = Revenue(x) - Cost(x)

  • Profit (x) = R(x) - C(x)

       Profit(x)=300x-\dfrac{x^2}{20}-\big(72,000+70x\big)

       Profit(x)=230x-\dfrac{x^2}{20}-72,000\\\\\\Profit(x)=-\dfrac{x^2}{20}+230x-72,000

ii) Find the first derivative and equal to 0 (it will be a maximum because the quadratic function is a parabola that opens downward)

  • Profit' (x) = -x/10 + 230
  • -x/10 + 230 = 0
  • -x + 2,300 = 0
  • x = 2,300

Thus, the production level that will realize the maximum profit is 2,300 units.

iii) Find the maximum profit.

You must substitute x = 2,300 into the equation for the profit:

  • Profit(2,300) = - (2,300)²/20 + 230(2,300) - 72,000 = 192,500

Hence, the maximum profit is $192,500

iv) Find the price the company should charge for each television set:

Use the price-demand equation:

  • p(x) = 300 - x/20
  • p(2,300) = 300 - 2,300 / 20
  • p(2,300) = 185

Therefore, the company should charge a price os $185 for every television set.

<em>3. ​Part (C) If the government decides to tax the company ​$4 for each set it​ produces, how many sets should the company manufacture each month to maximize its​ profit? What is the maximum​ profit? What should the company charge for each​ set?</em>

i) Now you must subtract the $4  tax for each television set, this is 4x from the profit equation.

The new profit equation will be:

  • Profit(x) = -x² / 20 + 230x - 4x - 72,000

  • Profit(x) = -x² / 20 + 226x - 72,000

ii) Find the first derivative and make it equal to 0:

  • Profit'(x) = -x/10 + 226 = 0
  • -x/10 + 226 = 0
  • -x + 2,260 = 0
  • x = 2,260

Then, the new maximum profit is reached when the production level is 2,260 units.

iii) Find the maximum profit by substituting x = 2,260 into the profit equation:

  • Profit (2,260) = -(2,260)² / 20 + 226(2,260) - 72,000
  • Profit (2,260) = 183,800

Hence, the maximum profit, if the government decides to tax the company $4 for each set it produces would be $183,800

iv) Find the price the company should charge for each set.

Substitute the number of units, 2,260, into the equation for the price:

  • p(2,260) = 300 - 2,260/20
  • p(2,260) = 187.

That is, the company should charge $187 per television set.

7 0
3 years ago
PLEASE PLEASE PLEASE HELP
bekas [8.4K]

Answer:

its the last choice

Step-by-step explanation:

3 0
4 years ago
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How would you represent a situation that has more than 1 solution? What does the solution represent about that situation? Explai
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Answer:

You don't

Step-by-step explanation:

For starters there is only one answer to every math question if you get an answer make sure it is correct

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. Find the simple interest on a $2,219.00 principal, deposited for 6 years at a
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SI=PTR/100

SI=2129*6*1.91/100

SI=24398.34/100

SI=243.9834$

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