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julia-pushkina [17]
3 years ago
8

Help pls

Mathematics
1 answer:
ki77a [65]3 years ago
3 0

Answer:

Distance between the school and the theater = 2\sqrt{13} units

Step-by-step explanation:

Use distance formula to calculate the distance from the school (8, 7) and the theater (2, 3):

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

school (8, 7) = (x_1, y_1)

theater (2, 3) = (x_2, y_2)

Plug in the values into the distance formula

d = \sqrt{(2 - 8)^2 + (3 - 7)^2}

d = \sqrt{(-6)^2 + (-4)^2}

d = \sqrt{36 + 16} = \sqrt{52}

Simplify

d = \sqrt{4 \times 13}

d = 2\sqrt{13}

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exis [7]

Answer: x - 777 = 0

Step-by-step explanation

x = 777

Move the contrast to the left

x - 777 = 777 -777

Eliminate the opposites

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4 0
3 years ago
I have corner points of:
WARRIOR [948]
The points you found are the vertices of the feasible region. I agree with the first three points you got. However, the last point should be (25/11, 35/11). This point is at the of the intersection of the two lines 8x-y = 15 and 3x+y = 10

So the four vertex points are:
(1,9)
(1,7)
(3,9)
(25/11, 35/11)

Plug each of those points, one at a time, into the objective function z = 7x+2y. The goal is to find the largest value of z

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

Plug in (x,y) = (1,9)
z = 7x+2y
z = 7(1)+2(9)
z = 7+18
z = 25
We'll use this value later. 
So let's call it A. Let A = 25

Plug in (x,y) = (1,7)
z = 7x+2y
z = 7(1)+2(7)
z = 7+14
z = 21
Call this value B = 21 so we can refer to it later

Plug in (x,y) = (3,9)
z = 7x+2y
z = 7(3)+2(9)
z = 21+18
z = 39
Let C = 39 so we can use it later

Finally, plug in (x,y) = (25/11, 35/11)
z = 7x+2y
z = 7(25/11)+2(35/11)
z = 175/11 + 70/11
z = 245/11
z = 22.2727 which is approximate
Let D = 22.2727

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

In summary, we found
A = 25
B = 21
C = 39
D = 22.2727

The value C = 39 is the largest of the four results. This value corresponded to (x,y) = (3,9)

Therefore the max value of z is z = 39 and it happens when (x,y) = (3,9)

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

Final Answer: 39

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Answer:

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