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romanna [79]
3 years ago
8

The distance from a focus to a vertex on the graph of a parabola is not the same as the distance from vertex to a perpendicular

point on the directrix.
True or false?
Mathematics
1 answer:
harkovskaia [24]3 years ago
3 0
Hey there,

Your question is stating: <span>The distance from a focus to a vertex on the graph of a parabola is not the same as the distance from vertex to a perpendicular point on the directrix. 

True or false?

This would be a </span>TRUE statement. <span>The distance from a focus to a vertex on the graph of a parabola is not the same as the distance from vertex to a perpendicular point on the directrix would be </span>TRUE.

Hope this helps.

~Jurgen

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Assume each figure shown has the same orientation. Which figure is the image of square LMNP after a translation of
umka21 [38]

Figure 4 is the image of the square LMNP after the translation.

<u>Step-by-step explanation:</u>

Let us see the coordinates of the pre image LMNP as,

L (-3,1)

M(-1,1)

N(-1,-1)

P(-3,-1)

after translation of (x,y) → (x+5, y -3) the coordinates of the image obtained as,

L'(2,-2)

M'(4,-2)

N'(4,-4)

P'(2,-4) which matches the image 4.

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Does anyone know how to do this?
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Each guest would get 6 party favors.

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First, we find out how many party favors Amy bought. 3 bags of 12 party favors equals 36 party favors (3x12)

So, divide 36 party favors by 6 guests, and you get the final answer of 6 party favors per guest.

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6 0
3 years ago
Better Products, Inc., manufactures three products on two machines. In a typical week, 40 hours are available on each machine. T
Kaylis [27]

Answer:

z (max)  =  1250 $

x₁  = 25    x₂  =  0   x₃  =  25

Step-by-step explanation:

                                Profit $    mach. 1      mach. 2

Product 1     ( x₁ )       30             0.5              1

Product 2    ( x₂ )       50             2                  1

Product 3    ( x₃ )       20             0.75             0.5

Machinne 1 require  2 operators

Machine   2 require  1  operator

Amaximum of  100 hours of labor available

Then Objective Function:

z  =  30*x₁  +  50*x₂  +  20*x₃      to maximize

Constraints:

1.-Machine 1 hours available  40

In machine 1    L-H  we will need

0.5*x₁  +  2*x₂  + 0.75*x₃  ≤  40

2.-Machine 2   hours available  40

1*x₁  +  1*x₂   + 0.5*x₃   ≤  40

3.-Labor-hours available   100

Machine 1     2*( 0.5*x₁ +  2*x₂  +  0.75*x₃ )

Machine  2       x₁   +   x₂   +  0.5*x₃  

Total labor-hours   :  

2*x₁  +  5*x₂  +  2*x₃  ≤  100

4.- Production requirement:

x₁  ≤  0.5 *( x₁ +  x₂  +  x₃ )     or   0.5*x₁  -  0.5*x₂  -  0.5*x₃  ≤ 0

5.-Production requirement:

x₃  ≥  0,2 * ( x₁  +  x₂   +  x₃ )  or    -0.2*x₁  - 0.2*x₂ + 0.8*x₃   ≥  0

General constraints:

x₁  ≥   0       x₂    ≥   0       x₃     ≥   0           all integers

The model is:

z  =  30*x₁  +  50*x₂  +  20*x₃      to maximize

Subject to:

0.5*x₁  +  2*x₂  + 0.75*x₃  ≤  40

1*x₁  +  1*x₂   + 0.5*x₃       ≤  40

2*x₁  +  5*x₂  +  2*x₃        ≤  100

0.5*x₁  -  0.5*x₂  -  0.5*x₃  ≤ 0

-0.2*x₁  - 0.2*x₂ + 0.8*x₃   ≥  0

x₁  ≥   0       x₂    ≥   0       x₃     ≥   0           all integers

After 6 iterations with the help of the on-line solver AtomZmaths we find

z (max)  =  1250 $

x₁  = 25    x₂  =  0   x₃  =  25

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3 years ago
Find an equation for the perpendicular bisector of the line segment whose endpoints are (8,-5)(8,−5) and (4,3)(4,3).
Marysya12 [62]

Answer:

Correct answer\\y=\frac{1}{2} x-4

Step-by-step explanation:

y-\frac{-5+3}{2} =\frac{-1}{\frac{3-(-5)}{4-8} } (x-\frac{4+8}{2} )\\y+1=\frac{1}{2} (x-6)\\y=\frac{1}{2} x-4

6 0
3 years ago
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