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alukav5142 [94]
3 years ago
9

Linear Programming:

Mathematics
1 answer:
tatuchka [14]3 years ago
7 0
Let the number of large bookcases be x and number of small bookcases be y, then
Maximise P = 80x + 50y;
subkect to:
6x + 2y ≤ 24
x, y ≥ 2

The corner points are (2, 2), (2, 6), (3.333, 2)
For (2, 2): P = 80(2) + 50(2) = 160 + 100 = 260
For (2, 6): P = 80(2) + 50(6) = 160 + 300 = 460
For (3.333, 2): P = 80(3.333) + 50(2) = 266.67 + 100 = 366.67

Therefore, for maximum profit, he should produce 2 large bookcases and 6 small bookcases.
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What is the answer to 17=5k-2
nadezda [96]


17=5k-2

[add 2 on both sides]

19=5k

[divide 5 on both sides]

k=19/5(3.8)


HOPE THIS HELPS!!!!!!

6 0
3 years ago
Read 2 more answers
Hey Siri the sum of two numbers is 50 and the larger number is six less than 3 times the smaller number find the smaller number
Inga [223]

x + y = 50

x = 3y - 6


3y - 6 + y = 50

4y = 56

y = 14


8 0
3 years ago
A real estate company wants to build a parking lot along the side of one of its buildings using 800 feet of fence. If the side a
Elden [556K]

Answer:

80,00ft^{2}

Step-by-step explanation:

According to my research, the formula for the Area of a rectangle is the following,

A = L*W

Where

  1. A is the Area
  2. L is the length
  3. W is the width

Since the building wall is acting as one side length of the rectangle. We are left with 1 length and 2 width sides. To maximize the Area of the parking lot we will need to equally divide the 800 ft of fencing between the <u>Length and Width.</u>

800 / 2 = 400ft

So We have 400 ft for the length and 400 ft for the width. Since the width has 2 sides we need to divide 60 by 2.

400/2 = 200 ft

Now we can calculate the maximum Area using the values above.

A = 400ft*200ft

A = 80,000ft^{2}

So the Maximum area we are able to create with 800 ft of fencing is 80,00ft^{2}

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

Read more on Brainly.com - brainly.com/question/12953427#readmore

3 0
3 years ago
Hey!
noname [10]

Answer:

A. Cylinder + cone

<u>Volume is the sum of volumes:</u>

  • V = Vcon + Vcyl = 1/3πr²h₁ + πr²h₂
  • V = 1/3π*9²*12 + π*9²*120 = 31554.2 cm³

<u>Surface area of cone:</u>

  • A = A=πr(r+√(h₁²+r²)) = π*9(9 + √(9²+12²)) = 678.6 cm²

<u>Surface area of cylinder minus bases:</u>

  • A = 2πrh₂ = 2π*9*120 = 6785.8 cm²

<u>Total surface area:</u>

  • 678.6 + 6785.8 = 7464.4 cm²

-------------------------------------------------

B. Cube+ pyramid

<u>Volume:</u>

  • V = a³ + (1/3)a²h = a³ + (1/3)a²√(l²-(a/2)²)
  • V = 8³ + (1/3)8²√(10²-4²) = 707.5 cm³

<u>Surface area of pyramid:</u>

  • A = a² + 2al = 8² + 2*8*10 = 224 cm²

<u>Surface area of cube minus bases:</u>

  • A = 4a² = 4(8²) = 256 cm²

<u>Total surface area:</u>

  • A = 224 + 256 = 480 cm²
5 0
3 years ago
What are the numbers to them?
Alex Ar [27]

Answer:

I believe for top last number is 49

and bottom first empty is 11 and second empty is 19

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
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