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anygoal [31]
3 years ago
13

-x+4(1+5x)=156please help me with this

Mathematics
2 answers:
Ostrovityanka [42]3 years ago
6 0

Answer:

The correct answer is x = 8

Step-by-step explanation:

-x + 4 + 20x = 156

-x + 20x = 19x

19x + 4 = 156

156 - 4 = 152

19x = 156

x = 156/9

x = 8


<u>Check my answer:</u>

-8 + 4 + 20(8) = 156

-4 + 160 = 156

SpyIntel [72]3 years ago
3 0
-x + 4 + 20x = 156
19x + 4 = 156
- 4. -4
19x = 152
19. 19
X = 8





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1.) -2(6 - 2x) = 4(-3 + x)<br> 2.) 5 - 1(2x + 3) = -2(4 + x)
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Let's solve your equation step-by-step.

Question 1:   −2(6−2x) =4(−3+x)

 

Step 1: Simplify both sides of the equation.

−2(6−2x) =4(−3+x)

(−2) (6) +(−2) (−2x) =(4)(−3)+(4)(x)(Distribute)

−12+4x=−12+4x

4x−12=4x−12

 

Step 2: Subtract 4x from both sides.

4x−12−4x=4x−12−4x

−12=−12

 

Step 3: Add 12 to both sides.

−12+12=−12+12

0=0

Answer: All real numbers are solutions.

Question 2:  

Let's solve your equation step-by-step.

5−1(2x+3) =−2(4+x)

 

Step 1: Simplify both sides of the equation.

5−1(2x+3) =−2(4+x)

5+(−1) (2x) +(−1) (3) =(−2) (4)+(−2)(x)(Distribute)

5+−2x+−3=−8+−2x

(−2x) +(5+−3) =−2x−8(Combine Like Terms)

−2x+2=−2x−8

−2x+2=−2x−8

 

Step 2: Add 2x to both sides.

−2x+2+2x=−2x−8+2x

2=−8

 

Step 3: Subtract 2 from both sides.

2−2=−8−2

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Answer: There are no solutions.

 

8 0
3 years ago
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6 0
2 years ago
Find m AEB<br> 66.5<br> 133<br> 266<br> 227
Sergeu [11.5K]

Answer:

The measure of arc AEB is 266\°

Step-by-step explanation:

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3 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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