<p>Can someone explain the following to me?</p>
<p>3) <a href="http://i252.photobucket.com/albums/hh17/xxEliza321xx/Photoon8-28-11at152PM.jpg%5B/url%5D">http://i252.photobucket.com/albums/hh17/xxEliza321xx/Photoon8-28-11at152PM.jpg</a>
They are asking for the area of ABC (AC=4)</p>
<p>8) If (rs^2)(t^3)(u^3)=0, which of the following must be true?</p>
<p>Narrowed it down to:
a) rt<0
e) st>0</p>
<p>I can't seem to figure out why the answer is A, but not E. </p>
<p>Thanks.</p>
<p>I tried doing the first one. No luck :(</p>
<p>Would be interested in knowing the method. Anyone?</p>
<p>I think the only thing you are missing is angle ACB. You should be able to do 180 - 60 and find that it is 120. Then you have 1 side and all the angles of triangle ABC; thus you can use law of sines to get the rest of the stuff you need. You can use law of sines to find BC (4, or you can just see it would be the same as AC); then you can use the law of sines to get BD (or you could do CD if you wanted to); then, with two sides of triangle BCD, use Pythagorean theorem to get the last side. Once you have all that, you can find that area easily.</p>
<p>The 2nd one makes no sense to me. These are probably harder than what you will see on the Math 2 test.</p>
<p>The first one you just have to find the angle of the second triangle which is 120 and since the other two sides are 30 each its isosoleces and two sides are 4. Which means 4 is the hypotenouse of the 1st triangle(one to the left). Since that one is a 30-60-90 and its hypotenouse is 4, its base is 2 and height is 2sqrt3. Now you need to subtract area of the entire triangle 6(2sqrt3) = 12sqrt3/2 - area of left triangle 2 times 2sqrt3 = 4sqrt3/2 =</p>
<p>6 sqrt3 - 2 sqrt 3 = 4sqrt 3</p>
<p>Im working on the second one and ill post when I find the method. Honestly it makes no sense cause you know one of them must be zero and the rest could be any value. Or my logic is screwed up.</p>
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<p>Can you post the exact problem statement and exact list of answers, please?
There must be some restriction on the values of r, s, t, and u. If not, could set all to zero to satisfy equation.</p>
<p>Modification to number 8:</p>
<p>u<0</p>
<p>All of the choices were:</p>
<p>a) rt<0
b) urt<0
c) r>0
d) st>0
e) st>0</p>
<p>Clearly, I had SOME understanding of this problem, but why is the answer A and not E?</p>
<p>Is this problem 8 of test 1 section 3 in Dr. Chung’s SAT Math book? </p>
<p>Then your problem was mistyped. Here is the corrected statement of the problem.</p>
<p>If (rs^2)(t^3)(u^3) > 0 and u < 0, which of the following must be true?</p>
<p>a) rt<0
b) urt<0
c) r>0
d) st>0
e) st>0</p>
<p>Solution: Since u < 0 , u^3 < 0.
This implies that (rs^2)(t^3) is also < 0 since (rs^2)(t^3)(u^3) > 0
Therefore, (rs^2)(t^3) = rt * s^2 * t^2 < 0.
s^2 * t ^2 > 0 since they are squares
This implies that rt < 0 since rt * s^2 * t^2 < 0</p>
<p>^ Hey, you really “know the stuff,” don’tcha? But yeah, your problem and reasoning makes a lot of sense!</p>
<p>Ok, but doesn’t using all negative numbers disprove this though? I used r=-5, s=-2, t=-3, and u=-1…maybe I made a careless mistake somewhere?</p>
<p>The answer is a) rt<0. If r and t were both negative, then the correct answer wouldn’t be true. (-5)(-3) = 15, which isn’t < 0.</p>
<p>Oh okay. Then it was careless then. I must have forgotten that it has to be less than 0 when I took negatives into account. Thanks :)</p>
