Need Math help from ACt practice test 2010!

<ol>
<li>The sum of an infinite geometric series with the first term A and a common ratio,1 is given by A/1-r. The sum of a given infinite geometric series is 200, and the common ratio is 0.15. What is the second term of this series?</li>
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<p>The other ones I need help with are numbers, 48. 49, 50, and 53 but they are too hard to post. </p>

<p>I really need help! thanks</p>

<p>heres the link for the test:</p>

<p><a href="http://www.act.org/aap/pdf/preparing.pdf%5B/url%5D"&gt;http://www.act.org/aap/pdf/preparing.pdf&lt;/a&gt;&lt;/p>

<p>Sum of the series is 200, and the common ratio is 0.25. SO:</p>

<p>200 = A/ (1-0.15) = A/ (0.85)</p>

<p>So, A = 200(0.85) = 170</p>

<p>Now A = 170, which is the first term. Since r = 0.15, the second term is A * r, or 170 * 0.15, which is 25.5</p>

<ol>
<li>You see the angle in between the B side and the sqrt(a^2 + b^2) side? Let that angle be X. Then:</li>
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<p>tanX = a/b</p>

<p>now lets say we take the tan-1 (tan inverse) of both sides. Then:</p>

<p>tan-1(tanX) = tan-1(a/b)</p>

<p>tan-1 and tan cancel, so you get:</p>

<p>X = tan-1(a/b)</p>

<p>The question asks what is the cos (tan-1(a/b)), which is the same as asking what is cosX. It’s easy to see that cosX is b/sqrt(a^2 + b^2), so the answer is D.</p>

<ol>
<li><p>Just by looking, you know that K is wrong. Even functions are symmetric about the y-axis, and just plugging in a few numbers makes it obvious that F is the answer. This is not a hard question - review even/odd functions.</p></li>
<li><p>82x + 1 = 41 − x</p></li>
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<p>8^(2x+1) = 4^(1-x)</p>

<p>Get both in the same base, which is 2.</p>

<p>2^3(2x+1) = 2^2(1-x)</p>

<p>Since they are the same base, set the exponents equal to each other.</p>

<p>3(2x+1) = 2(1-x)
6x + 3 = 2 - 2x
8x = -1
x = -1/ 8</p>

<p>C</p>

<ol>
<li>F</li>
</ol>

<p>It explains how to graph complex numbers. Basically, a is the distance down the x axis, and b is the distance down the y axis, and the hypotenuse of that “triangle” is the modulus. The point with the biggest modulus is the one farthest away from the origin, which is z1. Again, this question is not hard, but you need to read thoroughly.</p>

<p>Thanks. Now that I see it, you are. It isn’t hard, I just haven’t had that math in so long and forgot it!</p>