Annoying me I’ve been at this for hours
1 answer:
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Using it's vertex, the maximum value of the quadratic function is -3.19.
<h3>What is the vertex of a quadratic equation?</h3>A quadratic equation is modeled by:
The vertex is given by:
In which:
Considering the coefficient a, we have that:
- If a < 0, the vertex is a maximum point.
- If a > 0, the vertex is a minimum point.
In this problem, the equation is:
y + 4 = -x² + 1.8x
In standard format:
y = -x² + 1.8x - 4.
The coefficients are a = -1 < 0, b = 1.8, c = -4, hence the maximum value is:
More can be learned about the vertex of a quadratic function at brainly.com/question/24737967
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Answer:
The answer is "Options A, B, and E represent mutually exclusive events".
Step-by-step explanation:
Two occurrences that can happen immediately called mutually incompatible. Let's now glance at our options and figure out where the statements are mutually incompatible events.
In Option A: You could see that landing on an unwanted portion and arriving on 2 are events that are locally incompatible, even as undesirable portion contains 3 and 4, and 2 were shaded.
In Option B: Arriving on a shaded part and falling on 3 are also mutually incompatible because there are 3 on a windows azure.
In Option C: A darkened portion and an increasing amount can land while 2 would be an even number as well as on the shaded portion. That number is very much the same.
In Option D: At the same time as 4 is greater than 3 and it is situated upon an undistressed section, landing and attracting a number larger than 3 can happen.
In Option E: Landing on a shaded part and landing on even a shaded part is an excluding event, since shaders may either be shaded or unlit.
Answers:
Vertical asymptote: x = 0
Horizontal asymptote: None
Slant asymptote: (1/3)x - 4
<u>Explanation:</u>
d(x) =
=
Discontinuities: (terms that cancel out from numerator and denominator):
Nothing cancels so there are NO discontinuities.
Vertical asymptote (denominator cannot equal zero):
3x ≠ 0
<u>÷3</u> <u>÷3 </u>
x ≠ 0
So asymptote is to be drawn at x = 0
Horizontal asymptote (evaluate degree of numerator and denominator):
degree of numerator (2) > degree of denominator (1)
so there is NO horizontal asymptote but slant (oblique) must be calculated.
Slant (Oblique) Asymptote (divide numerator by denominator):
- <u>(1/3)x - 4 </u>
- 3x) x² - 12x + 20
- <u>x² </u>
- -12x
- <u>-12x </u>
- 20 (stop! because there is no "x")
So, slant asymptote is to be drawn at (1/3)x - 4