PLEASE HELP!!!! 100 ÷ 25 • 20 + 4 ÷ 2 – 14
2 answers:
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Answer:
Choice B is correct
Step-by-step explanation:
The eccentricity of the conic section is given as 4 and thus the conic section is a hyperbola. Hyperbolas are the only conic sections with an eccentricity greater than 1.
Next, the directrix of this hyperbola is located at y = -6 implying that the hyperbola will be opening upwards. Consequently, the polar equation of this hyperbola will be of the form;
The value of k in the numerator is the product of eccentricity and the absolute value of the directrix;
k= 4*6 = 24
The polar equation is thus given by alternative B
See below for the demonstration of (e) graphing a line using x- and y-intercepts
<h3>The graphing method to demonstrate</h3>To solve this question, we make demonstrate (e) graphing a line using x- and y-intercepts
<h3>The learning goals</h3>At the end of this exercise, you should be able to graph a line using x- and y-intercepts
<h3>A brief overview</h3>A linear equation is represented as:
Ax + By = C
Where:
- y-intercept = C/B
- x-intercept = C/A
This means that we plot the points (0,C/B) and (C/A, 0) on a graph and then connect the points using a straight line.
Once this is done, the line has been graphed.
For instance, a line parallel to 3x + 8y = 16 will have a slope of -3/8
<h3>An original problem to solve</h3>Plot the graph of the following equation using the intercepts
2x + 4y = 10
We have:
2x + 4y = 10
Set x = 0 to determine the y-intercept
4y = 10
Divide by 4
y = 2.5
Set x = 0 to determine the x-intercept
2x = 10
Divide by 2
x = 5
So, we have:
(5, 0) and (0, 2.5)
Plot the above points and connect the points (see attachment for graph)
<h3>An original problem to solve by another student</h3>Plot the graph of the following equation using the intercepts
3x + 6y = 18
We have:
3x + 6y = 18
Set x = 0 to determine the y-intercept
6y = 18
Divide by 6
y = 3
Set x = 0 to determine the x-intercept
3x = 18
Divide by 3
x = 6
So, we have:
(6, 0) and (0, 3)
Plot the above points and connect the points (see attachment for graph)
Read more about linear equations at:
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