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3 years ago

HELPPP IM ALMOST DONE WITH THIS CLASS

Mathematics

1 answer:

iris [78.8K]3 years ago

5 0

<h3>Answer: Choice D) y = cos(2x)</h3>

Explanation:

Normally, cosine has a period of 2pi. This means that the curve repeats itself every 2pi units. However, this graph has a period of pi.

We can see this by noting that the distance from one peak such as x = 0 to another adjacent peak x = pi is exactly pi units across.

Since T = pi is the period, we then know that

B = (2pi)/T

B = (2pi)/(pi)

B = 2

Then recall that the general template is y = Acos(B(x-C))+D

In this case, A = 1, C = 0 and D = 0. So all of this leads to y = cos(2x)

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2 years ago

Which is the interval notation represent the domain?

Kazeer [188]

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The first choice, $(-\infty, 7) \cup (7, \infty)$ .

Step-by-step explanation:

The dashed vertical line is a vertical asymptote. The x-coordinate of all points on this vertical asymptote are equal to 7. In other words, when the x-value of a point on the graph approaches $7$ , its y-value approaches infinity (or negative infinity.)

Either way, the graph is not defined for $x = 7$ . The point $7$ should thus be excluded from the domain of the graph.

The graph is apparently defined for all other x-values. The domain of the function should thus be all real numbers with the exception of $x = 7$ . Here's how to write that in interval notation:

The set of all real numbers can be expressed as the interval $(-\infty, \infty)$ . Note that infinity $\infty$ (or $(- \infty)$ itself isn't a real number. The round brackets indicate that both endpoints are excluded from the from the interval.
The set of all real numbers less than (not equal to) $7$ is $(-\infty,7 )$ (both ends are excluded.) The set of all real numbers greater than (not equal to) $7$ is $(7, \infty)$ .
The set of all real numbers that is not equal to $7$ is the union of all real numbers less than $7$ and all real numbers greater than $7$ . The union operator $\cup$ connects two intervals.

In other words, the domain is $(-\infty, 7) \cup (7, \infty)$ .

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3 years ago