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Elanso [62]
3 years ago
14

Please I need this done quickly Explain the answer please

Mathematics
1 answer:
Mashutka [201]3 years ago
5 0
When writing an equation of a line, keep in mind that you ALWAYS need two pieces of information when you go to write an equation:
1. ANY point on the line
2. Slope (m)
3. Y-intercept (b)

Once you have these two pieces of information, you plug the x and y values from your point and the slope (m) value into the slope-intercept form, y = mx + b.


Start by solving for the slope of each line. We need two ordered pairs (x, y) from each line to solve for the slope:

Offer A:
(0, 400) and (80, 700)

Let (x1, y1) = (0, 400)
(x2, y2) = (80,700)

m = (y2 - y1)/(x2 - x1)
m = (700 – 400)/(80 – 0) = 300/80 = 15/4

Therefore, the slope (m) for Offer A = 15/4

Next, we need the y-intercept. The y-intercept is the y-coordinate of the point where the graph of the linear equation crosses the y-axis. The y-intercept is also the value of y when x = 0. One of the points we used for solving the slope for Offer A reflects the description for the y-intercept, which is point (0, 400). The y-coordinate is 400–this is the y-intercept of the line.

Now, we can establish the linear equation for Offer A: y = 15/4x + 400

Do the same steps for the other offer:

Offer B:
(0, 500)) and (100,900)
Let (x1, y1) = (0, 500)
(x2, y2) = (100,900)

m = (y2 - y1)/(x2 - x1)
m = (900 - 500)/(100 – 0) = 400/100 = 4

Therefore, the slope for Offer B is 4.

Next, we need the y-intercept. For Offer B, the line crosses at point (0, 500). The y-coordinate, 500 is the y-intercept (b) of the line.

The linear equation for Offer B is:
y = 4x + 500



**NOTE: please double-check the coordinates that I used for solving the slopes of each line before using my inputs. I had a slight difficulty going through the coordinates of your graph because it’s a bit blurry (using my phone to write this post).

Please mark my answers as the Brainliest if you find my explanations helpful :)
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Summary of even and odd functions

A function f is even if the graph of f is symmetric with respect to the y-axis. Algebraically, f is even if and only if f(-x) = f(x) for all x in the domain of f.

A function f is odd if the graph of f is symmetric with respect to the origin. Algebraically, f is odd if and only if f(-x) = -f(x) for all x in the domain of f.

From the definition itself, we see how can we test if the functions are even or odd.

- First graph (to the left)

From the graph, we see that the function is symmetric with respect to the y-axis because the left side (x<0) is a reflection (along the y-axis) of the right side (x>0), so it is an even function.

- Second graph (to the right)

From the graph, we see that the function is symmetric with respect to the origin, the left side (x<0) is a reflection (along with the origin) of the side to the right (x>0), so it is an odd function.

- First function (to the left)

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g(x)=5x^2

is an even function because we see that:

g(-x)=5\cdot(-x)^2=5x^2=g(x)

- Second function (to the right)

The function:

h(x)=-2x^5+3x^3

is an odd function because we see that:

h(-x)=-2\cdot(-x)^5+3\cdot(-x)^3=-(-2x^5+3x^3)=-h(x)^{}

Summary

We have the following results:

- First function (to the left): even

- Second graph (to the right): odd

- First function (to the left): even

- Second function (to the right): odd

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