Can someone help me immediately on this question having to do with triangle ABC?

Mathematics

1 answer:

SVETLANKA909090 [29]3 years ago

7 0

<h3>Answers:</h3><h3>a. Vertices of triangle ABC are: A, B, C</h3><h3>b. Sides of triangle ABC are: AB, BC, AC</h3><h3>c. The side between angle A and angle C is: side AC</h3><h3>d. The angle between sides AB and CA is: angle A</h3><h3>e. Scalene triangle</h3>

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Explanations:

a. Each uppercase letter represents a point or angle of the triangle.
b. Connect two points of a triangle and you get a line segment. The order of the letters does not matter. So AB is the same as BA.
c. Like with part b, connecting two angles or points forms a segment.
d. Note how the letter "A" is in both AB and CA, so this is the shared angle between the two segments.
e. Sides AB, BC, and AC are all different lengths, so we have a scalene triangle. If you had two sides equal to each other, then you'd have an isosceles triangle. If all three sides are equal, then it would be equilateral.

There is no need for a diagram, but if you want, you can draw one out. See the attached image below for the diagram. This diagram should hopefully answer any questions you may have about the explanations above. There are many ways to draw the triangle, so your diagram might look different from mine.

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Show all work to identify the asymptotes and state the end behavior of the function f(x)=5x/x-25

Nina [5.8K]

Using the asymptote concept, it is found that:

The vertical asymptote is of x = 25.

The horizontal asymptote is of y = 5.

Considering the horizontal asymptote, it is found that the end behavior of the function is that it tends to y = 5 to the left and to the right of the graph.

<h3>What are the asymptotes of a function f(x)?</h3>

The vertical asymptotes are the values of x which are outside the domain, which in a fraction are the zeroes of the denominator.

The horizontal asymptote is the value of f(x) as x goes to infinity, as long as this value is different of infinity.

In this problem, the function is:

$f(x) = \frac{5x}{x - 25}$

Considering the denominator, the vertical asymptote is:

x - 25 = 0 -> x = 25.

The horizontal asymptote is found as follows:

$y = \lim_{x \rightarrow \infty} f(x) = \lim_{x \rightarrow \infty} \frac{5x}{x - 25} = \lim_{x \rightarrow \infty} \frac{5x}{x} = \lim_{x \rightarrow \infty} 5 = 5$