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blondinia [14]
3 years ago
11

Linear programming find the minimum value

Mathematics
1 answer:
Paha777 [63]3 years ago
5 0

Answer:

Step-by-step explanation:

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Answer:

6,000= y

2hrs= x

y=x

6,000y= 2=x

Step-by-step explanation:

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Find the quotient for 7.868+1.4
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Step-by-step explanation:

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The length of a rectangle is 7 more than the width. The area is 744 sqaure yards, find the length and width of the rectangle​
AVprozaik [17]

Answer:

  • Length of rectangle is 31 yards and Width is 24 yards.

Given:

  • The length of a rectangle is 7 more than the width.
  • The area is 744 sqaure yards

Solution:

Let's assume Width of rectangle be x and Length of rectangle be x + 7 respectively.

Using formula

\\  \:  \:  \:  \:  \pink{ \dashrightarrow \:  \:  \:  \:  \sf { \underbrace{Area_{(Rectangle)} =  Length × Width  }}} \\  \\

On Substituting the required values, we get;

\\ \: \: \: \:  \dashrightarrow \: \: \: \: \sf (x)(x + 7) = 744 \\ \\ \\ \: \: \: \:  \dashrightarrow \: \: \: \: \sf  {x}^{2}  + 7x = 744 \\ \\ \\ \: \: \: \:  \dashrightarrow \: \: \: \: \sf  {x}^{2}  + 7x - 744 = 0 \\ \\ \\  \: \: \: \:  \dashrightarrow \: \: \: \: \sf  {x}^{2}  + 31x - 24x - 744 = 0 \\ \\ \\ \: \: \: \:  \dashrightarrow \: \: \: \: \sf x(x + 31) - 24 (x + 31) = 0 \\ \\ \\ \: \: \: \:  \dashrightarrow \: \: \: \: \sf (x + 31)(x - 24) = 0 \\ \\ \\ \: \: \: \:  \dashrightarrow \: \: \: \: \sf x = 24 \:  or \:   - 31 \\ \\

As we know that width of the rectangle can't be negative. So x = 24

<u>Hence</u>,

  • Width of rectangle = x = 24 yards
  • Length of the rectangle = x + 7 = 31 yards

\thereforeLength of rectangle is 31 yards<u> </u>and Width is 24 yards.

8 0
2 years ago
Whats the answer to number 9?
agasfer [191]

Answer:

B and D

Step-by-step explanation:

Because only those two equations on the graphs are straight. and are increasing or decreasing in a constant pace.

4 0
2 years ago
You have been tasked with filling 4 ounce and 3 ounce bags from a 41 ounce container of candy.
Alchen [17]

Answer:

A. 3 possible combinations

B. 8 4-ounce's bags and 3 3-ounce's bags

C. 2 4-ounce's bags and 11 3-ounce's bags

D. 8 4-ounce's bags and 3 3-ounce's bags

E. All solutions offer the same revenue.

Step-by-step explanation:

You have been tasked with filling 4 ounce and 3 ounce bags from a 41 ounce container of candy. Let x be the number of 4 ounce bags and y be the number of 3 ounce bags. Then

4x+3y=41.

A. Find all integer solutions:

  • When x=0, then 3y=41 - impossible, because 41 is not divisible by 3.
  • When x=1, then 3y=37 - impossible, because 37 is not divisible by 3.
  • When x=2, then 3y=33, y=11 - possible.
  • When x=3, then 3y=29 - impossible, because 29 is not divisible by 3.
  • When x=4, then 3y=25 - impossible, because 25 is not divisible by 3.
  • When x=5, then 3y=21, y=7 - possible.
  • When x=6, then 3y=17 - impossible, because 17 is not divisible by 3.
  • When x=7, then 3y=13 - impossible, because 13 is not divisible by 3.
  • When x=8, then 3y=9, y=3 - possible.
  • When x=9, then 3y=5 - impossible, because 5 is not divisible by 3.
  • When x=10, then 3y=1 - impossible, because 1 is not divisible by 3.

You get 3 possible combinations.

B. 1. 2 + 11 = 13,

2. 5 + 7 = 12,

3. 8 + 3 = 11.

The minimal number of bags is 11.

C. 1. 2·7+11·5=69 cents

2. 5·7+7·5=70 cents

3. 8·7+3·5=71 cents

The cheapest is 1st solution.

D. 1. 2·6+11·5=67 cents

2. 5·6+7·5=65 cents

3. 8·6+3·5=63 cents

The cheapest is 3rd solution.

E. 1. 2·2+11·1.50=$20.50

2. 5·2+7·1.50=$20.50

3. 8·2+3·1.50=$20.50

All solutions offer the same revenue.

5 0
3 years ago
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