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andrey2020 [161]
3 years ago
13

What does a flowchart contain of

Mathematics
1 answer:
icang [17]3 years ago
7 0
A flowchart can also be defined as a diagrammatic representation of an algorithm, a step-by-step approach to solving a task. The flowchart shows the steps as boxes of various kinds, and their order by connecting the boxes with arrows. This diagrammatic representation illustrates a solution model to a given problem.
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How to simplify 14 2/3<br> ​
Norma-Jean [14]

Answer:

You can look down

Step-by-step explanation:

Okay so you can put this has a mixed faction.

so it is 46/3. and then you can simplify it and you get the same answer. 14 2/3 but you still cant simplify it because all numbers are un-even. like you cant simplify.

8 0
2 years ago
Read 2 more answers
Please help. I’ll mark you as brainliest if correct!
Readme [11.4K]

Answer:

\large \boxed{\sf \ \ x=0, \ \ y=-5 \ \ }

Step-by-step explanation:

Hello, please consider the following.

We have two equations:

(1) -2x - 4y = 20

(2) -3x + 5y = -25

5*(1)+4*(2) gives

   -10x - 20y -12x + 20y = 100 - 100 = 0

   -22x = 0

   x = 0

I replace in (1)

   -4y = 20

   y = -20/4 = -5

There is one solution x = 0, y = -5

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

3 0
3 years ago
Read 2 more answers
Solve each. Show all necessary work.
LenaWriter [7]

Answer:

y = 3x - 9

Step-by-step explanation:

First, find the slope using y2 - y1 / x2 - x1

36 - 21 = 15

15 - 10 = 5

15/5 = 3, so the slope (m) is 3.

Now, plug into point-slope form: y - y1 = m (x - x1)

y - 21 = 3 (x - 10)

Simplify

y - 21 = 3x - 30

y = 3x - 9

I hope this helps!!

8 0
2 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
2 years ago
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It’s 56689 + the power of everything is 25 round it up
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