Math IIC Problems Inside - Help?

<p>Hi - I was taking the Barrons practice tests and there were some problems that just threw me off. If anyone can explain the reasoning behind getting some of the answers, I'd really appreciate it. Also, if some of there answers can be worked out on the TI-89 Titanium, I'd appreciate if you can explain which tools to use for that. I'm taking the test this Saturday.
Thanks :)</p>

<p>1) For all positive angles less than 360 degrees, if csc(2x+30 degrees)=cos(3y-15 degrees), the sum of x and y is
A)185 B) 65 C) 35 D) 215 E) 95 (all answer choices are in degree)</p>

<p>2) (2 cis 50 degrees)^3 written in rectangular form is
A)6.9+4i B) 4-6.9i C) 6.9-4i D)-6.9+4i E) -4+6.9i</p>

<p>3) In order for the inverse of f(x)=cos(3x) to be a function, the domain of f must be limited to
A) 0 <=x <= 60
B) 0 <=x <= 180
C) 90 <=x <= 270
D) 30 <=x <= 90
E) -30 <=x <= 30</p>

<p>4) An equation in polar form equivalent to x^2 + y^2 - 4x + 2 = 0 is
A)r=4cos(theta) + 2
B)r^2 = 4cos(theta) + 2
C)4r=cos(theta)
D)r^2-4rcos(theta)+2=0
E)r^2=4rcos(theta)</p>

<p>5) Which of the following is the equation of the circle that has its center at the origin and is tangent to the line with equation 3x-4y=10?
A)x^2+y^2=2
B)x^2+y^2=4
C)x^2+y^2=3
D)x^2+y^2=5
E)x^2+y^2=10</p>

<p>6) In how many ways can a committee of four be selected from nine men so as to always inlcude a particular man?
A)84 B)70 C)48 D) 56 E)126</p>

<p>7) A rectangular box has dimensions of length=6, width=4, and height=5. The angle formed by a diagonal of the box contains
A)27 degrees
B)35 degrees
C)40 degrees
D)44 degrees
E)55 degrees</p>

<p>Will put more up as I come across them. Thanks in advance.</p>

<p>what are the answers?</p>

<p>ok i just did number one, i might do the others...there might be another way...but this way was pretty fast...and used more logic than math..
csc(2x+30 degrees)=cos(3y-15 degrees)
1/(sin(2x+30)=cos(3y-15)
1=sin(2x+30)*cos(3y-15)
Now here, I was just like...which values of sin and cos when mutliplied together equal 1? well duh, 1 *1, and the sin(90)=1 and the cos(0)=1, so then i just solved 2x+30=90, and got 30, and sovled 3y-15=0, and got 5...added together and you get 35</p>

<ol>
<li> Just Plug in the answers, use the "(2nd) Entry" thing if your calculator has it, that will speed it up.</li>
<li>If you had a TI-83, I know going to the degree menu has a Polar to Rectangular thing</li>
<li>Dunno here, I would try graphing (arc cos(x))/3 and see what pops up.</li>
<li>Dunno, never come across any of that crap in any practice tests or the real thing. I think you can graph that in polar mode, not sure I don't have my calculator.</li>
<li>Graph y=(3x-10)/4 and then see what circle graph is tangent.</li>
<li>nCr thing?</li>
<li>sq rt((L^2+W^2)+H^2)= length of diagonal, arc sin or arc cos of height divided by diagonal will give the angles, or you could just use arc tan of the length of the bottom diagonal and the height.</li>
</ol>

<p>there's no way of plugging in the answers for number one...every answer is the sum of x and y and you dont know what those are...i guess you could keep guessing...lol...as far as i know...my way is the only way.</p>

<p>lol, I didn't read the whole question. I would probably skip that question. There is probably some lame trig identity that makes that question super easy.</p>

<p>Thanks :)
"6. nCr thing?"</p>

<p>Yup, but what do you do with it? It's not as simple as plugging in 9C4 :P</p>

<p>for six, i got 56...i'm not sure if my logic is correct here, because combinations aren't exactly my thing...but if 9 men are to make a 4 person comittee that always includes the same one man...wouldnt it just be 8 people making a 3-man committe? so i did (8x7x6)/(3x2x1), because it's a combination since order doesnt matter...i think that's right..but i could be wrong</p>

<p>mmkay, you're right - the answer is 56. I hadn't thought of it that way.
Thanks :)</p>

<p>for question one, -1*-1 is also a possibility. and they ask for all POSITIVE angles less than 360, so 0 degrees in cos 0 is not an option</p>

<p>therefore do the same except use 180 (cos is -1) and 270(sin is -1) as your angles and you will get 185 degrees</p>

<p>oh yeah...0's neither positive nor negative...anyway, for your information OP, barrons is harder than the real thing. dont let it fool you. trust me. barrons is always almost harder...princeton review/kaplan more close for math ii...sparknotes is a tad bit too easy, but still more realistic than barrons.</p>

<p>mmkay, I took the test this morning. Barrons defintely prepared me well for it. :D</p>