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STALIN [3.7K]
3 years ago
13

Subtract.

Mathematics
2 answers:
notka56 [123]3 years ago
6 0
-2 4/7 is your answer
Karo-lina-s [1.5K]3 years ago
3 0

Answer: The required simplified mixed number is -2\dfrac{4}{7}

Step-by-step explanation:

Since we have given that

-3\dfrac{2}{7}-(-\dfrac{5}{7})

We need to solve this.

First we change the mixed fraction into improper fraction.

3\dfrac{2}{7}=\dfrac{23}{7}

So, it becomes,

-\dfrac{23}{7}+\dfrac{5}{7}\\\\=\dfrac{-23+5}{7}\\\\=\dfrac{-18}{7}\\\\=-2\dfrac{4}{7}

Hence, the required simplified mixed number is -2\dfrac{4}{7}

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What are the solution(s) to the quadratic equation 40 - x2 = 0?
Zanzabum

Answer:

  x=\pm2\sqrt{10}

Step-by-step explanation:

You want to solve ...

  40 -x^2=0\\\\40=x^2\qquad\text{add $x^2$}\\\\\pm\sqrt{40}=x\qquad\text{take the square root}\\\\\boxed{x=\pm2\sqrt{10}}\qquad\text{simplify}

5 0
3 years ago
Helpppp<br>What is 25% of 307<br>10<br>|<br>2<br>12<br>1<br>7<br>2​
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Answer:

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Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
there are 9 boys in a class of 14 students. a)what is the ratio of boys to girls. b) what is the ratio of girls to all students
ElenaW [278]

Answer:

a). 9 to 5

b). 5 to 14

Step-by-step explanation:

If there are 9 boys then there are 5 girls because 14 - 9 = 5.

So, there are 5 girls of 14 students in the class.

4 0
3 years ago
Which ordered pair is a solution of the equation y equals x minus 3
Rom4ik [11]
Y = x-3

Basically you just need to plug in an x value and solve for y
y = x - 3
y = (4) - 3
y = 1

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Hope this helps 
6 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
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