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andrezito [222]
1 year ago
9

The x-intercepts of cosine become what for the secant function?

Mathematics
1 answer:
Anuta_ua [19.1K]1 year ago
8 0

using the definition of secant we know that,

sec(x)=\frac{1}{cos(x)}

then, the definition of x-intercept is that the function evaluated is equal to 0 then, if f(x) is equal to cos(x)

f(x)=cos(x)

we can say that if the function is equal to 0 then the function secant will look something like this

\begin{gathered} sec(x)=\frac{1}{0} \\  \end{gathered}

and we know that any number divided by 0 is undefined, so the x.intercepts make the function secant undefined, meaning that the x-intercepts become vertical asymptotes in the secant function.

Answer:

The x-intercepts of the function cosine become the vertical asymptotes of the secant function.

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Based on the data given, the conclusion that can be drawn from the correlation coefficient associated with the linear equation is C. There's a weak negative correlation between the variables.

<h3>How to illustrate the information?</h3>

From the information given, it can be seen that the value of y is increasing with the increase in the value of x.

This implies that the correlation is positive. Also, the change in the values of y with x are scattered a lot.

Therefore, the conclusion that can be drawn from the correlation coefficient associated with the linear equation is that there's a weak negative correlation between the variables

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Solve equation:<br> -3|d-5|=-6
gulaghasi [49]

Answer:

d = 7, 3

Step-by-step explanation:

First, you need to isolate the absolute value

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Sveta_85 [38]

I'll do Problem 8 to get you started

a = 4 and c = 7 are the two given sides

Use these values in the pythagorean theorem to find side b

a^2 + b^2 = c^2\\\\4^2 + b^2 = 7^2\\\\16 + b^2 = 49\\\\b^2 = 49 - 16\\\\b^2 = 33\\\\b = \sqrt{33}\\\\

With respect to reference angle A, we have:

  • opposite side = a = 4
  • adjacent side = b = \sqrt{33}
  • hypotenuse = c = 7

Now let's compute the 6 trig ratios for the angle A.

We'll start with the sine ratio which is opposite over hypotenuse.

\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(A) = \frac{a}{c}\\\\\sin(A) = \frac{4}{7}\\\\

Then cosine which is adjacent over hypotenuse

\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}\\\\\cos(A) = \frac{b}{c}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\

Tangent is the ratio of opposite over adjacent

\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(A) = \frac{a}{b}\\\\\tan(A) = \frac{4}{\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{\sqrt{33}*\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{(\sqrt{33})^2}\\\\\tan(A) = \frac{4\sqrt{33}}{33}\\\\

Rationalizing the denominator may be optional, so I would ask your teacher for clarification.

So far we've taken care of 3 trig functions. The remaining 3 are reciprocals of the ones mentioned so far.

  • cosecant, abbreviated as csc, is the reciprocal of sine
  • secant, abbreviated as sec, is the reciprocal of cosine
  • cotangent, abbreviated as cot, is the reciprocal of tangent

So we'll flip the fraction of each like so:

\csc(\text{angle}) = \frac{\text{hypotenuse}}{\text{opposite}} \ \text{ ... reciprocal of sine}\\\\\csc(A) = \frac{c}{a}\\\\\csc(A) = \frac{7}{4}\\\\\sec(\text{angle}) = \frac{\text{hypotenuse}}{\text{adjacent}} \ \text{ ... reciprocal of cosine}\\\\\sec(A) = \frac{c}{b}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(\text{angle}) = \frac{\text{adjacent}}{\text{opposite}} \ \text{  ... reciprocal of tangent}\\\\\cot(A) = \frac{b}{a}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\

------------------------------------------------------

Summary:

The missing side is b = \sqrt{33}

The 6 trig functions have these results

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Rationalizing the denominator may be optional, but I would ask your teacher to be sure.

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