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zlopas [31]
3 years ago
8

The SneakerRama Company makes and sells sneakers. They have one linear function that represents the cost of producing sneakers a

nd another linear function that models how much income they get from those sneakers. Describe the key features that would determine if these linear functions ever intercepted. (10 points)
Mathematics
1 answer:
maw [93]3 years ago
6 0

Answer:

key features are the slopes of both linear functions, if they are not equal then the two linear  functions will be intercepted

Step-by-step explanation:

Any linear function can be represented by the following linear equation:

y=mx+b

where m is the slope of function

and y in the intercept

Given:

The SneakerRama Company makes and sells sneakers.

They have one linear function that represents the cost of producing sneakers

Let it be represented by y₁=m₁x+b₁

and another linear function that models how much income they get from those sneakers

Let it be represented by y₂=m₂x+b₂

Now if m₁≠m₂

then the two functions will intercept at some point on the graph at some value of x. At that point the cost of producing sneakers will be equal to income they get from those sneakers.

y₂-y₁=0

This point is called the break-even point and the profit is zero at this point !

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Answer:

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Step-by-step explanation:

28/44 = F/99

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Read 2 more answers
Find the point on the parabola y^2 = 4x that is closest to the point (2, 8).
guapka [62]

Answer:

(4, 4)

Step-by-step explanation:

There are a couple of ways to go at this:

  1. Write an expression for the distance from a point on the parabola to the given point, then differentiate that and set the derivative to zero.
  2. Find the equation of a normal line to the parabola that goes through the given point.

1. The distance formula tells us for some point (x, y) on the parabola, the distance d satisfies ...

... d² = (x -2)² +(y -8)² . . . . . . . the y in this equation is a function of x

Differentiating with respect to x and setting dd/dx=0, we have ...

... 2d(dd/dx) = 0 = 2(x -2) +2(y -8)(dy/dx)

We can factor 2 from this to get

... 0 = x -2 +(y -8)(dy/dx)

Differentiating the parabola's equation, we find ...

... 2y(dy/dx) = 4

... dy/dx = 2/y

Substituting for x (=y²/4) and dy/dx into our derivative equation above, we get

... 0 = y²/4 -2 +(y -8)(2/y) = y²/4 -16/y

... 64 = y³ . . . . . . multiply by 4y, add 64

... 4 = y . . . . . . . . cube root

... y²/4 = 16/4 = x = 4

_____

2. The derivative above tells us the slope at point (x, y) on the parabola is ...

... dy/dx = 2/y

Then the slope of the normal line at that point is ...

... -1/(dy/dx) = -y/2

The normal line through the point (2, 8) will have equation (in point-slope form) ...

... y - 8 = (-y/2)(x -2)

Substituting for x using the equation of the parabola, we get

... y - 8 = (-y/2)(y²/4 -2)

Multiplying by 8 gives ...

... 8y -64 = -y³ +8y

... y³ = 64 . . . . subtract 8y, multiply by -1

... y = 4 . . . . . . cube root

... x = y²/4 = 4

The point on the parabola that is closest to the point (2, 8) is (4, 4).

4 0
3 years ago
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