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Mathematics

1 answer:

levacccp [35]3 years ago

8 0

For the given triangle, x = 5.1 cm and y = 6.1 cm.

Step-by-step explanation:

Step 1:

In the given triangle, the angle is 50°. It is given that the opposite side has a length of y cm and the adjacent side has a length of y cm. The hypotenuse of the triangle measures 8 cm.

To determine the length of the opposite side of the triangle, we use the sin of the given angle.

To determine the length of the adjacent side of the triangle, we use the cos of the given angle.

$sin \theta = \frac{oppositeside}{hypotenuse} , cos \theta = \frac{adjacentside}{hypotenuse}.$

Step 2:

In the given triangle,

The length of the opposite side = y cm,

The length of the adjacent side = x cm,

The length of the hypotenuse = 8 cm,

The angle of the triangle = 50°.

$sin \theta = \frac{oppositeside}{hypotenuse} , sin 50 = \frac{y}{8}.$

$y = 0.7660 (8) = 6.128.$

$cos \theta = \frac{adjacentside}{hypotenuse} , cos 50 = \frac{x}{8}.$

$x = 0.6427 (8) = 5.1416.$

So x = 5.1416 cm and y = 6.128 cm. Rounding these off to one decimal place, we get x = 5.1 cm and y = 6.1 cm.

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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:

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