<p>It thought it was very straight forward.</p>
<p>For the last two,
49) 2Sqrt2 (whatever that simplifies to)
50) 2 solutions?</p>
<p>Can’t remember what the question was for number 49.</p>
<p>For 50, I think you’re right. For the equation to be 1, either the exponent had to be zero, or the base had to be 1. The exponent had two real zeros, but the base only had imaginary zeros. So yeah, two.</p>
<p>i said infinitely many solutions because it’s either when the exponent is equal to zero (which had two solutions) or when the base = 1…once the base = 1 there are infinitely many things it can be raised to to equal one</p>
<p>For 49, it was a pic of a circle and it asked for the length of AD. You had to use a 306090 triangle.</p>
<p>How many can you miss to get a 780? a 760? Do you all know from prior tests?</p>
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<p>I don’t think there was any way that the base could equal 1. IIRC, I set the base equal to one, minus’d 1 from both sides and used the quadratic formula. No real zeros, just imaginary.</p>
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<p>Angle A = 120º. BA was the side opening from the 30º angle, and since the side opening up from the 60º angle was 4, the answer was sqrt 3/4 I think.</p>
<p>wait sorry this sounds really dumb but whats the difference between real and imaginary numbers? arent imaginary numbers the ones with “i” in them? idk i used a TI 89 to solve when the equation was equal to zero and i got numbers that didn’t have “i” in them</p>
<p>I graphed the exxponent 1 and it had 2 solutions.</p>
<p>Can you draw a pic of 49.</p>
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<p>I think you got this question right. i put four solutions myself, because teh base was a quadratic opening upwards with a negative y-intercept, so it definitely can equal 1…so I found the two zeros and the two 1s, but failed to realize that with it has base 1 it has infinately many solutions.</p>
<p>Too bad for me i guess…I did better on Math 1 on practice tests, but for some reason I feel better about math 2.</p>
<p>for 49 there was the side opp the 30deg angle side, and it said it was 4…so that means that the other side is 4(sqrt3), because it was a 30-60-90 triangle</p>
<p>Oops I just noticed that I wrote 2Sqrt2.
I first got 4 as my answer, but when I checked it, I did somethikn wrong. Can I get a pic of the drawing if ne1 remebers it so i can try 2 solve.</p>
<p>There never seems to be enough time for CB’s math sections. I ran out of time without starting #48 and left three or four other questions blank.</p>
<p>was the bottom equation x^2-x-6? or x^2+x-6?</p>
<p>^ it does not matter its still opening upwards with a negative y-intercept</p>
<p>i did quadratic and i got 2 answers for the bottom, unfoiled the exponent and got 2 solutions there too, so i put 4</p>
<p>It can’t be infiinate because 1^(infinate) would mean the x values change, making the base different than 1. </p>
<p>I really think it was 4.
(2 values to make the base=1 and 2 to make the power=0)</p>
<p>i put 4 as well b/c when you unfoil your exponent becomes x^4, meaning that it will cross 4 times</p>
<p>I graphed it and it had 2 solutions. Do you guys remember the whole thing?</p>
<p>the other 2 are imaginary solutions</p>
<p>looks like not many took this test.</p>
<p>does anyone remember F(x) = (x-3) g(x)</p>
<p>was the answer zero</p>