The set of inequalities y ≤ x² - 3 and y > -x² + 2 do not have a solution
<h3>How to modify the graphs</h3>
The attached figure 1 represents the missing piece in the question
From the graph, we have:
f(x) = x² - 3
g(x) = -x² + 2
Next, we change the equations to inequalities as follows:
y ≤ x² - 3
y > -x² + 2
To modify the graph, we then perform the following transformations:
- Shift the function g(x) down by 2 units
- Reflect across the x-axis
- Shift the function g(x) down by 3 units
<h3>How to identify the solution set</h3>
After the modifications in (a), we have:
y ≤ x² - 3 and y > -x² + 2
Next, we plot the graph of the inequalities
From the graph of the inequalities, the curves of the inequalities have no point of intersection
Hence, the set of inequalities do not have a solution
Read more about inequalities at:
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Total divided by number of rows=number per row
63/7=9
9 per row
$10.5
Your car would cost a total of $25.5 to get to Detroit. and your friends car would cost $15.5. So you end up saving a total amount of $10.50
Hello :
1 ) the 1 quadrant : the sine, cosine and tangent are <span> positives
2) the 2 </span> quadrant : the sine is positive but cosine and tangent are negatives
3 ) the 3 quadrant : the tangent is positive but cosine and sine are negatives
4 ) the 4 quadrant : the cosine is positive but tangent and sine are negatives
Answer:
Vertical asymptote at x= 6
Oblique asymptote at y=x+10
Step-by-step explanation:
To find out vertical asymptote , we take the denominator and set it =0
x-6=0 , so x=6
Vertical asymptote at x= 6
The degree of numerator is 2 and the degree of denominator is 1.
Here, the degree of numerator is higher than the degree of denominator. So there will be a slant or oblique asymptote. we can find it by long division.
x + 10
-------------------------------
x-6 x^2 + 4x -3
x^2 - 6x
-------------------------------(subtract)
10x - 3
10x - 60
-----------------------------(subtract)
57
Oblique asymptote at y=x+10
