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Cloud [144]
2 years ago
9

How many fixed-requirement constraints does a transportation problem with 5 factories and 6 customers have?a. 30.b. 11.c. 6.d. 5

.
Mathematics
1 answer:
bulgar [2K]2 years ago
7 0

Answer:

Option B

Step-by-step explanation:

From the question we are told that:

Demand point m=5

Supply Point n=6

Generally the equation for fixed-requirement constraints is mathematically given by

 X=m+n

 X=5+6

 X=11

Therefore the correct option is

Option B

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Step-by-step explanation:

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7x^2=9+x what are the values of x<br><br>Will give medal and points
DENIUS [597]
7x² = 9 + x   Subtract x from both sides
7x² - x = 9    Subtract 9 from both sides
7x² - x - 9 = 0   Use the Quadratic Formula

a = 7 , b = -1 , c = -9

x = \frac{-b \pm  \sqrt{b^2 - 4ac} }{2a}   Plug in the a, b, and c values
x = \frac{- (-1) \pm  \sqrt{(-1)^2 - 4(7)(-9)} }{2(7)}   Cancel out the double negative
x = \frac{1 \pm  \sqrt{(-1)^2 - 4(7)(-9)} }{2(7)}   Square -1
x = \frac{1 \pm  \sqrt{1 - 4(7)(-9)} }{2(7)}   Multiply 7 and -9
x = \frac{1 \pm  \sqrt{1 - 4(-63} }{2(7)}   Multiply -4 and -63
x = \frac{1 \pm  \sqrt{1 + 252} }{2(7)}   Multiply 2 and 7
x = \frac{1 \pm  \sqrt{1 + 252} }{14}   Add 1 and 252
x = \frac{1 \pm  \sqrt{253} }{14}   Split up the \pm
x = \left \{ {{ \frac{1 +  \sqrt{253} }{14} } \atop { \frac{1 -  \sqrt{253} }{14} }} \right.
The approximate square root of 253 is <span>15.905973.
</span>x ≈ \left \{ { \frac{1 + 15.905973}{14} } \atop { \frac{1 - 15.905973}{14} }} \right   Add and subtract
x ≈ \left \{ {{ \frac{16.905973}{14} } \atop { \frac{14.905973}{14} }} \right.   Divide
x ≈ \left \{ {{1.2075} \atop {1.0647}} \right.   Round to the nearest hundredth
x ≈ \left \{ {{1.21} \atop {1.06}} \right.

<span>
</span>
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