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Hitman42 [59]
3 years ago
11

How would you describe domain

Mathematics
1 answer:
irina [24]3 years ago
6 0

Answer:

The domain represents the x-axis, more specifically, what is happening on the x-axis. So when looking at a graph, if you are asked to find the domain think about what the x-axis looks like. I put an image in to show you an example of what the domain would be for a parabola:

So on the left side of the x-axis, we can see that the line stretches out into negative infinity, so the domain would begin at negative infinity.

On the right side of the x-axis, the parabola also stretches into positive infinity, so here the domain would be (negative infinity, positive infinity), because it goes to both ends forever.

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A swimmer swims 4 laps in 3 minuets. What is the swimmers unit rate?
AURORKA [14]
Hello there! Your answer to this question is (4/3 laps per min) hope it helps and have a wonderful day!
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3 years ago
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28. What is the maximum number of real roots
attashe74 [19]
The maximum number of real roots is 5 because the degree is 5. The fundamental theorem of algebra states that the degree of a polynomial is the number of real solutions. (The degree is the highest exponent of the polynomial)
6 0
3 years ago
Derivatives concept:
mr Goodwill [35]

9514 1404 393

Answer:

  A. y = -2x +13

  B. y = 8x -7

Step-by-step explanation:

A. We can read the y-intercept of the secant line from the graph. It is 13.

The slope can also be read from the graph, but we choose to use the slope formula:

  m = (y2 -y1)/(x2 -x1)

  m = (19 -9)/(-3 -2) = 10/-5 = -2

Then the slope-intercept formula for the line is ...

  y = mx + b . . . . . . line of slope m and y-intercept b

  y = -2x +13

__

B. The vertex of the given parabola is (0, 1). We notice that when x=1 (1 unit right of the vertex), y = 3 (2 units up from the vertex). This tells us the vertical scale factor of the parabola is 2. That means the vertex form equation is ...

  y = a(x -h)^2 +k . . . . . . . . vertex (h, k), scale factor 'a'

  y = 2(x-0)^2 +1 . . . . . . . use known values for (h, k)

  y = 2x^2 +1

The derivative of this is ...

  y' = 4x

So, at x=2, the given point A, the slope of the tangent line is ...

  m = y' = 4(2) = 8

We have a point and the slope, so we can write the point-slope form of the equation for the tangent line:

  y -9 = 8(x -2)

Rearranging to slope-intercept form, this is ...

  y = 8x -7

__

<em>Additional comment</em>

You can also read the slope of the tangent line from the graph. The line also goes through the point (1, 1), so has a rise of 8 for a run of 1. The y-intercept can be found from ...

  b = y -mx = 9 -8(2) = -7

This lets you write the equation of the tangent line directly from the graph.

That is, the parameters of both lines can be read from the graph, so there is very little "development" required.

8 0
3 years ago
I need to know how many cupcakes are needed to fill this box.
Gre4nikov [31]

Is it more to this question like numbers or something I don’t really understand
4 0
4 years ago
Add the equations.<br>2x-3y = -1<br>+ 3x + 3y= 26<br><br>​
hichkok12 [17]

Answer:

The answer in the procedure

Step-by-step explanation:

we have

2x-3y=-1 ----> equation A

3x+3y=26 --> equation B

Solve the system by elimination

Adds equation A and equation B

2x-3y=-1

3x+3y=26

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

2x+3x=-1+26

5x=25

x=25/5

x=5

Find the value of y

substitute the value of x in equation A or equation B and solve for y

2(5)-3y=-1

10-3y=-1

3y=10+1

y=11/3

8 0
3 years ago
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