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Lapatulllka [165]
3 years ago
12

(-6, -2) and (5, 1) find the distance between these two points using the distance formula

Mathematics
2 answers:
Paha777 [63]3 years ago
6 0

Answer:

The answer is 11.401754

Setler79 [48]3 years ago
6 0

Answer: Approximately 11.4 units

================================================

Work Shown:

(x1,y1) is the first point (-6,-2)

(x2,y2) is the second point (5,1)

Apply the distance formula

d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(-6-5)^2+(-2-1)^2}\\\\d = \sqrt{(-11)^2+(-3)^2}\\\\d = \sqrt{121+9}\\\\d = \sqrt{130}\\\\d \approx 11.4017542509913\\\\d \approx 11.4\\\\

The distance between the two points is roughly 11.4 units.

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Answer:

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Step-by-step explanation:

For the given situation let the number of tv sets be x and number of dvd players be y.

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it gives subject to the constraints

4x+2y≥100......(1)(100 hours each week. It takes 4 hours to refurbish a tv set and 2 hours to refurbish a dvd player.)

x+y≥35.......(2)(weekly sales target is to refurbish at least 35 tv sets or dvd players.)

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put x=0 gives 2y=100⇒y=50

put y=0 gives 4x=100⇒x=25

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Similarly for equation (2)

put x=0 gives y=35

put y=0 gives x=35

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with the help of (3) and (4) we plot the following graph (assume x≥0 and y≥0)

The unbounded feasible region determined by constraints gives the corner points as A(0,50),B(15,20)and C(35,0).

from  we get the value of z is minimum at point B (15,20) .

hence we got our optimal solution at B (15,20), where x is the number of tv sets and y is the number of dvd players  .therefore Ruhana and her team should refurbish 15 tv sets and 20 dvd players each week to minimize the cost.

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3 years ago
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Step-by-step explanation:

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Solnce55 [7]

Answer:

Step-by-step explanation:

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