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Papessa [141]
3 years ago
6

You have taken a long position in a call option on IBM common stock. The option has an exercise price of $142 and IBM's stock cu

rrently trades at $148. The option premium is $7 per contract. a. How much of the option premium is due to intrinsic value versus time value? b. What is your net profit on the option if IBM’s stock price increases to $158 at expiration of the option and you exercise the option? c. What is your net profit if IBM’s stock price decreases to $138?
Mathematics
1 answer:
REY [17]3 years ago
5 0

Answer:

Step-by-step explanation:

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State the domain of the relation ? ( there is more than one?) select all that apply
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Step-by-step explanation:

Domain: all x-values that are to be used (independent values).

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PLS ANSWER ASAP 30 POINTS!!! CHECK PHOTO! WILL MARK BRAINLIEST TO WHO ANSWERS
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

\sin(A) = \frac{4}{7}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\\tan(A) = \frac{4}{\sqrt{33}} = \frac{4\sqrt{33}}{33}\\\\\csc(A) = \frac{7}{4}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\

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

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