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aleksandrvk [35]
4 years ago
6

A linear application has the following constraints:20 y 20Which of the following is not a corner point of the feasible region?O

A) (0,8)O B) (3,5,0)O D) (5, 30

Mathematics
1 answer:
Schach [20]4 years ago
7 0
The answer is D)  (5,3)

The corner points of the feasible region for the given linear application are the following:
(0,8)
(0,0)
(3.5,0)
(5,3)

These points can be obtained by graphing the inequalities and locating the points of intersection. Or one inequality can be paired with another and the solution solved. Since there are 4 inequalities, there must be 4 corner points.
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the class marks of a distribution are 37 42 and 47 the class limits corresponding to class mark 42 are
I am Lyosha [343]

Answer:

39.5,44.5

Step-by-step explanation:

Let\ the\ lower\ limit\ and\ the\ upper\ limit\ of\ the\ first\ interval\ be\ u_1,v_1.\\Let\ the\ lower\ limit\ and\ the\ upper\ limit\ of\ the\ second\ interval\ be\ u_2,v_2.\\Let\ the\ lower\ limit\ and\ the\ upper\ limit\ of\ the\ third\ interval\ be\ u_3,v_3.\\Hence,\\As\ the\ class\ marks\ are\ uniform\ with\ an\ a\ difference\ of\ 5, the\\ observation\ is\ continuous\ and\ the\ class\ size\ is\ 5\ too.\\Hence,\\v_1=u_2, v_2=u_3\\Now,\\Lets\ consider\ the\ first\ two\ classes.\\37=\frac{u_1+v_1}{2} \\42=\frac{u_2+v_2}{2}\\By\ adding\ the\ equations:\\79= \frac{u_1+v_1}{2}+\frac{u_2+v_2}{2}\\79=\frac{u_1+v_1+u_2+v_2}{2}\\79=\frac{(u_1+v_2)+(v_1+u_2)}{2}\\79=\frac{(u_1+v_2)}{2}+\frac{(v_1+u_2)}{2}\\Now,\\As\ the\ mid-point\ between\ two\ points\ can\ be\ calculated\ by\\ its\ average:\frac{u+v}{2} \\Hence,\\u_2\ lies\ in\ the\ mid-point\ of\ u_1\ and\ v_2.\\u_2=\frac{(u_1+v_2)}{2}\\

Hence,\\By\ substituting\ u_2=\frac{(u_1+v_2)}{2},\\79=u_2+\frac{v_1+u_2}{2}\\As\ v_1=u_2[Proven],\\79=u_2+ \frac{u_2+u_2}{2}\\79=u_2+\frac{2u_2}{2}\\79=2u_2\\Hence,\\u_2=\frac{79}{2}\\u_2=39.5\\Now,\ as\ we\ already\ know\ that\ the\ class\ size=5,\\v_2-u_2=5\\Hence,\\v_2=5+u_2\\Here,\\v_2=5+39.5\\v_2=44.5\\Hence,\\The\ upper\ limit\ of\ the\ second\ interval=44.5\\The\ lower\ limit\ of\ the\ second\ interval=39.5\\

7 0
3 years ago
Which equation satisfies all three pairs of a and b values listed in the table?
Tom [10]

Answer:

y = 3x -10

Step-by-step explanation:

The function seems to increase at a constant rate of 3 units. This means that the equation that satisfies these points could be that of a straight line.

We use the first two points to find the slope of the line.

(0, -10)

(1, -7)

(2, -4)

The equation of a line is:

y = mx + b

Where m is the slope of the line and b is the cutoff point with the y axis.

To find the slope of a line we use the following equation:

m = \frac{y_2-y_1}{x_2-x_1}\\\\m = \frac{-7 - (- 10)}{1-0}

m = 3

So:

y = 3x + b

The cut point (b) is found by replacing in the previous equation, any of the three points provided and clearing b.

-7 = 3(1) + b

b = -10

Now we can write the equation of the line sought.

y = 3x -10

You can verify that the three points provided belong to this equation.

7 0
4 years ago
Read 2 more answers
What is the y-intercept of the graph of the function f(x) = x2 3x 5?.
Fiesta28 [93]

Answer:

f(x) =x2 3x 5?

c = 5

therefore the solpe = 5

3 0
2 years ago
6000 whole number multiplied by a power of ten
Firdavs [7]
6000=6×10^3 (3 zeros)
8 0
3 years ago
Read 2 more answers
The function f(x) is shown in the graph
user100 [1]

Answer:

the function is an exponential funtion.

Step-by-step explanation:

learned it

3 0
2 years ago
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