CPhill

5. The sin is < 0 in quadrants 3 and 4. And the cos < 0 in quadrants 2 and 3. But the sin and the cos  would have to have the same sign for the tangent to be positive. Thus, theta is a 3rd quadrant angle. And since the tan(theta)= 1 at 45 degrees, this is a 45 degree angle in quadrant 3. And the cos of 45 degrees  = 1/√2, or √2/2, if you prefer. But, since the cos is negative in this quadrant, then cos(theta) = - 1/√2 = -√2/2.

6.  

2 cos x = -1   divide both sides by 2

cos x = -1/2    ....    this occurs at (2/3)pi in the interval [0, pi)  and at -(2/3)pi in the interval [-pi, 0]. The second answer may be a little hard to see. Note that cos x = -1/2 at (4/3)pi. But, subtracting 2*pi from this gives us -(2/3)pi.......

The arcsin of (9/16) isn't 55.8. The arccos of (9/16) will give you that answer. Go to the onsite calculator. Hit the "2nd" button and you will see the "acos" button. Hit this, key in 9/16, hit "Enter" or the "=" key and you will get your result.

I'm wondering if the phrasing of this question should be "at least two."

Consider the following pentagon in which the vertices represent 5 people at a gathering........call them A,B, C,D, E.

Now, suppose that all the people there are total strangers to one another. Then, it's clear that there are more than two people who have the same number of friends at the gathering. In fact, there are 5 people who have the same number of friends.......namely ......none!!!

But, if the phrase includes"at least two," then it's true. Suppose we have some polygon with a "large number" of alphabetical vertices - call the number "n" -, and let each vertex represent a "person."  Now, connect every vertex with every other vertex and let these connections represent the associations between people. In this instance everyone knows the same number of people - (n-1). Now, erase any one of the connected lines. Then, the two vertexes (people) that were connected by this line now know the same number of people (n-2). And if we continued in this fashion of erasing lines, there would always be at least two vertices that had the same valence (valence is the number of lines meeting at one vertex). And thus, there would always be at least two people who knew the same number of people!! For instance, suppose we erased AB  and AC. Then B and C would know the same number of people - in this case, everybody else except "A." Or, if we erased AB, BC and AC, then A, B, and C would know the same number of people - (n-3).

I realize that this isn't a "formal" proof, but I think it demonstrates the idea. Perhaps someone on the site could give a more "formal" one??? I'm sure there's a simpler way, I just don't happen to know it....!!!

Go ahead, ND.......I'm rich enough now!!!

I've got the answer, ND...the man was the lighthousekeeper....when he died, no one was left to tend the light, and the ships crashed on the rocks !!!

We know an angle, the hypoteneuse, and we're looking for the opposite side.  So we have:

sin(65) = opp / 20     Multiplyimg both sides by 20, we have

20*sin(65)  = opp  ≈ 18.13 ft.

To convert to radians, multiply the degree  measure by  (pi/180)..so we have.....

112.5 * (pi/180) = (5.8) pi

I'm not well-versed in Physics, but I thimk I can do 1 and 3 for you.

1)  The first part of the trip took 6mi/(30 m/h) = 1/5 hr. And the second part of the trip took 6 mi / (60m/h)= 1/10hr.   So the total time was   (1/5 + 1/10) hr  = 15/50 hr  = 3/10 hr. And the average rate = D / Total time = 12mi / (3/10)hr  = 120/3 mph = 40 mph.

3) We have the height at any time is given by

h = -4.9t^2 + 14t        ...... and we're assuming that the object's intial position is h = 0.....so we have

5 = -4.9t^2 + 14t    and rearranging, we have

-4.9t^2 + 14t -  5 = 0

And using the on-site calculator to solve this, we have

$${\mathtt{\,-\,}}{\mathtt{4.9}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{14}}{\mathtt{\,\times\,}}{\mathtt{t}}{\mathtt{\,-\,}}{\mathtt{5}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = {\mathtt{\,-\,}}{\frac{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,-\,}}{\mathtt{10}}\right)}{{\mathtt{7}}}}\\
{\mathtt{t}} = {\frac{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{10}}\right)}{{\mathtt{7}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = {\mathtt{0.418\: \!418\: \!884\: \!019\: \!217\: \!8}}\\
{\mathtt{t}} = {\mathtt{2.438\: \!723\: \!973\: \!123\: \!639\: \!3}}\\
\end{array} \right\}$$

The object is at the 5m level on the descent at about 2.4 sec.

(4x + 3)^2 / 2 = 36     multiply both sides by 2

(4x + 3)^2  = 72         take the square root of both sides

4x + 3 = ±√72            subtract 3 from both sides

4x = ±√72 - 3                divide by 4 on both sides

x = [ ±√72 - 3] / 4

And since you specify that x > 0, then the solution is x = [ √72 - 3] / 4 = [ 6√2 - 3] / 4

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