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Likurg_2 [28]
3 years ago
13

I will definitely rate 5 stars/brainliest!!! HELP PLEASE!!! State University must purchase 1,100 computers from three vendors. V

endor 1 charges $500 per computer plus a delivery charge of $5,000. Vendor 2 charges $350 per computer plus a delivery charge of $4,000. Vendor 3 charges $250 per computer plus a delivery charge of $6,000. Vendor 1 will sell the university at most 500 computers; vendor 2, at most 900; and vendor 3, at most 400. Formulate an IP to minimize the cost of purchasing the needed computers (Please solve the LP with Lindo or Excel).
Engineering
1 answer:
romanna [79]3 years ago
3 0
Why 1+12+ Y3 < 1100
Says the state of university Need to purchase 1100 computers in total, we have the following answer on the way top
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motikmotik

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brake fade. loss of brake effectiveness due to overheating.

Explanation:

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1 year ago
The no-slip condition is:________
viva [34]

Answer:

The correct answer is option (c) An experimental observation that the velocity of a fluid in contact with a solid surface is equal to the velocity of the surface.

Explanation

Solution:

When a fluid is in proximity to the boundary the solid and the velocities are the same or uniform for the fluid and the surface, no slip condition does not exist.

However, because the no-slip meets the expectations for gas and liquids, this condition no way connected in this case of two solid in proximity.

hence, the other options are wrong here.

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3 years ago
Alternating current lesson 4 exam
zloy xaker [14]
Huh? whats the question?
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3 years ago
A heating system must maintain the interior of a building at 20°C during a period when the outside air temperature is 5°C and th
Anettt [7]

Answer:

a. W = 51,194.54 kJ

b. W = 102,390 kJ

c. W = 153,585 kJ

Explanation:

(COP)_{HP} =\frac{Desired-effectx}{Work-done}= \frac{Q_{1} }{W} \\\\(COP)_{HP} =(COP)_{Ideal}\\\\\frac{Q1}{W} =\frac{T_{1} }{T_{1} -T_{2} }

W=Q_{1} \frac{T_{1}-T_{2}  }{T_{1} }

a. the ground at 15°C.

T_{1}=20°C = 273 K + 20 = 293 K

T_{2}=15°C = 273 K + 15 = 288 K

Q_{1}=3x10^{6} kJ

W=3x10^{6} kJ \frac{293 K-288 K}{293 K}=3x10^{6} kJ \frac{5 K}{293 K}=3x10^{6} kJ x 0.017065}

W = 0.051195x10^{6} kJ

W = 51,194.54 kJ

b. a pond at 10°C.

T_{2}=10°C = 273 K + 10 = 283 K

W=3x10^{6} kJ \frac{293 K-283 K}{293 K}=3x10^{6} kJ \frac{10 K}{293 K}=3x10^{6} kJ x 0.034130}

W = 0.102390x10^{6} kJ

W = 102,390 kJ

c. the outside air at 5°C.

T_{2}=5°C = 273 K + 5 = 278 K

W=3x10^{6} kJ \frac{293 K-278 K}{293 K}=3x10^{6} kJ \frac{15 K}{293 K}=3x10^{6} kJ x 0.051195}

W = 0.153585x10^{6} kJ

W = 153,585 kJ

Hope this helps!

3 0
3 years ago
The maximum capacity of a 2-lane carriageway of a four lane dual carriageway is 2000 veh/hour. due to pipe laying operations the
Yuki888 [10]

Based on the maximum capacity as a result of the pipe-laying operations, and the maximum capacity without obstruction to the four-lane dual carriageway, the mean speed of traffic in the bottleneck is -2.73km.

<h3>What is the mean speed of traffic?</h3>

This can be found as:

= (Maximum restricted capacity - freeflowing rate) / (Kb - Ka)

= (1,100 - 1,500) / (209 - 62.5)

= -2.73 km/h

The rate that the queue outside the bottleneck grows is:

= Free flowing rate - (mean speed of traffic x ka)

= 1,500 - (-2.73 x 62.5)

= 1,670 veh/hour

Find out more on bottleneck at brainly.com/question/9551615

#SPJ1

3 0
1 year ago
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