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ddd [48]
3 years ago
7

The constraints of a problem are listed below. What are the vertices of the feasible region?​

Mathematics
2 answers:
Mamont248 [21]3 years ago
8 0

Answer:

Option 4 : (0.\frac{3}{2} ) \ , \ (0,2) \ , \ (6,0) \ , \ (\frac{9}{4} ,0)

Step-by-step explanation:

<u>See the attached figure:</u>

To find the vertices of the feasible region, we need to graph the constraints, then find the area included by them, then calculate the vertices which is the intersection between each two of them.

As shown, the shaded area represents the solution of the constraints

So, the vertices of the feasible region are:

(0.\frac{3}{2} ) \ , \ (0,2) \ , \ (6,0) \ , \ (\frac{9}{4} ,0)

Inessa05 [86]3 years ago
5 0

Answer:

d

Step-by-step explanation:

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Given a sequence is defined by the explicit definition LaTeX: t_n=\:n^2+nt n = n 2 + n, find the 4th term of the sequence. (ie L
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Answer:

t_{4}=20

Step-by-step explanation:

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What is k if : k!+48=48((k+1)^m)
lys-0071 [83]

Answer:

The value of k is greater than or equal to 0, i.e. k≥7.

Step-by-step explanation:

The given equation is

k!+48=48((k+1)^m)

The value of k must be a positive integer because k! is defined for k≥0, where k∈Z.

Subtract 48 from both the sides.

k!=48((k+1)^m)-48

k!=48((k+1)^m-1)

k!=48(k+1-1)(\frac{(k+1)^m-1}{(k+1)-1})

Using [\frac{r^m-1}{r-1}=r^{m-1}+r^{m-2}+...+1], we get

k!=48k((k+1)^{m-1}+(k+1)^{m-2}+...+1)

Divide both sides by 48k.

\frac{k!}{48k}=(k+1)^{m-1}+(k+1)^{m-2}+...+1

\frac{k(k-1)!}{48k}=(k+1)^{m-1}+(k+1)^{m-2}+...+1

\frac{(k-1)!}{48}=(k+1)^{m-1}+(k+1)^{m-2}+...+1

Note: The value of m can be 0 or 1.

The value of k is positive integer, so the right hand side of the above equation must be a positive integer.

Since RHS of the equation is positive integer, therefore (k-1)! is completely divisible by 48.

k-1\geq 6

Add 1 on both sides.

k\geq 6+1

k\geq 7

Therefore the value of k is greater than or equal to 0.

4 0
4 years ago
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