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blagie [28]
3 years ago
8

Explain the difference between linear and exponential models. How can you determine if a function is linear or exponential from

a table,
graph, and equation?
Mathematics
1 answer:
oksano4ka [1.4K]3 years ago
4 0
In linear functions, rate of change is constant: as x goes up, y will go up a consistent amount. In exponential functions, the rate of change increases by a consistent multiplier—it will never be the same, but there will be a pattern.
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4 The width of a rectangular dance floor
Brilliant_brown [7]

Answer:

\boxed{\sf a. \ 6x + 5}

Given:

f(x) = -2x + 5

To Find:

f(-3x)

Step-by-step explanation:

\sf Evaluate \ f(x) \ by \ substituting \ in \ the \\ \sf value \ of \ x = -3x \ into \ f(x): \\ \sf f( - 3x) = -2( - 3x) + 5 \\ \\ \ \ \ \ \ \ \ \ \ \ \ \  \sf =  6x + 5

6 0
4 years ago
Is x+y+1=0 a tangent of both y^2=4x and x^2=4y parabolas?
Lubov Fominskaja [6]

Answer:

  yes

Step-by-step explanation:

The line intersects each parabola in one point, so is tangent to both.

__

For the first parabola, the point of intersection is ...

  y^2 = 4(-y-1)

  y^2 +4y +4 = 0

  (y+2)^2 = 0

  y = -2 . . . . . . . . one solution only

  x = -(-2)-1 = 1

The point of intersection is (1, -2).

__

For the second parabola, the equation is the same, but with x and y interchanged:

  x^2 = 4(-x-1)

  (x +2)^2 = 0

  x = -2, y = 1 . . . . . one point of intersection only

___

If the line is not parallel to the axis of symmetry, it is tangent if there is only one point of intersection. Here the line x+y+1=0 is tangent to both y^2=4x and x^2=4y.

_____

Another way to consider this is to look at the two parabolas as mirror images of each other across the line y=x. The given line is perpendicular to that line of reflection, so if it is tangent to one parabola, it is tangent to both.

7 0
3 years ago
How to solve for the x
navik [9.2K]
I think its 15\52 im guessing 
8 0
4 years ago
Thank you so much if you answer
andreev551 [17]
Answer: B, 31/40
altenate form 0.775

4 0
3 years ago
Read 2 more answers
Solve the equation by finding the common multiple.
Monica [59]

Answer:

1. 17/24

2. 1 1/8

3. 1 1/6

4. 1 5/24

5. 1. 1/4

6 0
3 years ago
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