<p>Not quite sure what I'm doing wrong in #18, I'm pretty sure it's $38 but apparently it's $35. So yeah, please help me with the 4 problems (pic above) guys. Help would really be appreciated!</p>
<h1>58: Visualize the cross-section as a rectangle with a truncated right triangle. The cross-section area is 75*10 - (1/2)(50)(6) = 600 ft^2. (600 ft^2)(50 ft) = 30000 ft^3, J.</h1>
<h1>59: “Slide” the segment of length 2 so that the endpoint is at the origin. You will see that the y-intercept of k is 2sqrt(2). y = x + 2sqrt(2), B.</h1>
<h1>60: Minimize x(x-6) or x^2 - 6x. The “vertex” of the quadratic occurs at x = 6/2 = 3, so the minimum value of x(x-6) is 3(3-6) = -9, G.</h1>
<p>@rspence Thanks for your help! Looks like I got #18 right. As for #40, I actually got 84/85 on my own, but I didn’t see it as an answer choice so I didn’t answer the question. Why do you have to multiply 84/85 by 3?</p>
<p>Gotcha. ACT test makers are weird. Anyways, thanks for your help on the questions, I really appreciate it! I’m trying to get 34+ on the math portion. I’ve completed AP Calc BC but still get 32/33s due to dumb mistakes or questions like these =/</p>
<p>I didn’t answer them simply b/c I didn’t see the forum (it wasn’t bumped or anything).</p>
<ol>
<li><p>Helps if you know a bit of trig, but not required. What is the period of that function?</p></li>
<li><p>“x-7 is a factor” implies that x = 7 is a root of the equation 2x^2 - 11x + k = 0. Plug in x = 7, solve for k.</p></li>
<li><p>Draw a radius connecting the center of the circle with the edge of the top level of the support. You have a right triangle with hypotenuse 4 and side length 2. Find the other length. Remember to multiply by 2!</p></li>
<li><p>To optimize surface area, suppose the 8" x 4" is the face touching the ground. The area of the rectangular region is 80 sq feet and the “area” of each tile is (2/3 ft)(1/3 ft) = 2/9 sq ft. 80 divided by 2/9 is 360, B. It can be shown that you can tile the 8’ x 10’ area with 8" x 4" tiles that are all oriented the same way.</p></li>
</ol>
<ol>
<li><p>I always thought a period was when a line goes all the way up, down, and on its way up, the point where it intersects x axis is the period? I just completed Calc -_-</p></li>
<li><p>Ahh, -21. </p></li>
<li><p>How didn’t I see this earlier.</p></li>
<li><p>I did exactly that. Except instead of 1/3 and 2/3 feet, I converted everything into inches, which is why it wasn’t working out. Ughhh.</p></li>
</ol>
<ol>
<li><p>Not necessarily, the period is just how often it repeats. For example, it starts at (0,-1) and then repeats at (pi, -1) so you could say that the period is pi.</p></li>
<li><p>Converting to inches work as well, but it’s more likely to make an arithmetic error, especially if you’re not using a calculator.</p></li>
</ol>
<p>Oh nevermind! I got it, wow that was dumb how I didn’t see it before. All I had to do was plug in a number and see what the output was, then match it to the answers. </p>
<p>I used 2, so the coordinates were (2,3) and (4,1). Distance = ~2.8, which is D.</p>
<p>Would’ve gotten 35 if I had plugged in any value on the actual practice test -_-…</p>
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