Help with these Math IIC questions?

Test Preparation SAT Subject Test Preparation
1

<p>These are from Barron's, but I didn't find the explanations for these questions satisfactory, or I just didn't get it.</p>

<p>Now, there are a lot of questions so I don't expect you to help me with all of them, but any amount would be greatly appreciated</p>

<p>Remove the space between . and com in the links to see the pictures</p>

<p><a href="http://i.imgur"&gt;http://i.imgur&lt;/a>. com/zVNpPSF.jpg
Answer: D</p>

<p>How is choice D a reflection across the line y = x?</p>

<p><a href="http://i.imgur"&gt;http://i.imgur&lt;/a>. com/bpSjR7F.jpg
Answer: B</p>

<p>When you graph this on the calculator, the function stops at y = 1 and comes back down. So how is y > 0?</p>

<p>Now please explain the following questions, because I just don't get how to do them:
<a href="http://i.imgur"&gt;http://i.imgur&lt;/a>. com/M5MmDgl.jpg
<a href="http://i.imgur"&gt;http://i.imgur&lt;/a>. com/iBSShdy.jpg
<a href="http://i.imgur"&gt;http://i.imgur&lt;/a>. com/9p7xrBp.jpg
<a href="http://i.imgur"&gt;http://i.imgur&lt;/a>. com/YGv4YUi.jpg
<a href="http://i.imgur"&gt;http://i.imgur&lt;/a>. com/U3NZIbk.jpg
<a href="http://i.imgur"&gt;http://i.imgur&lt;/a>. com/bJw8Zh0.jpg
<a href="http://i.imgur"&gt;http://i.imgur&lt;/a>. com/NJbdXF8.jpg
<a href="http://i.imgur"&gt;http://i.imgur&lt;/a>. com/Vg23xAl.jpg
<a href="http://i.imgur"&gt;http://i.imgur&lt;/a>. com/g6fJ04w.jpg
Answers:
38. B
39. C
42. B
44. B
45. D
48. B
10. D
14. D
18. E
21. B</p>

2

<p>zVNpPSF.jpg: To find the inverse, just swap the x and y co-ordinates. D is clearly the graph redrawn with x and y co-ords switched.</p>

<p>bpSjR7F.jpg: Not entirely sure about this (don’t have a graphing calc) but y = 1/x^(2/3) seems like it would tend to infinity for small values of |x| and tend towards 0 for large values of |x|. x on the denominator also means x != 0 so y !=0. So I’d guess y > 0.</p>

<p>M5MmDgl.jpg:
38: Not sure I understand what he question is asking. Sounds to me like you can discount TTT as a possibility. HHT, HTH and THH satisfy the condition so I’d guess (B) 3/7.</p>

<p>39: It’s © (-0.29, 0.43, -0.86) right? I would guess that the magnitude is ~1 (can be verified with superpythagorean theorem) and it’s in the exact opposite direction (still parallel though).</p>

<p>iBSShdy.jpg:
42: Quick guess is (B). Ignoring gender, there are 15 people, thus 15C5 = 3003 possible committees. There are 6C3*9C2 = 720 possible committees with 3 men and 2 women. Therefore the answer is 720/3003 = 240/1001.</p>

<p>9p7xrBp.jpg:
44: I’m getting (B) t-1, but I’m not sure at all. If that’s correct I can show you my workings but more or less it’s guesswork.</p>

<p>Edit: since it IS correct, all I did was isolate y and equate the rest to the given parameter. Then I isolated x in the resulting equation. Result was (t-1).</p>

<p>YGv4YUi.jpg:
45: Plug in (x-3) instead of x, expand. Unless I made a careless mistake, it’s ax^2 + (b-6a)x + 9a - 3b +c. Axis of symmetry would be x = (6a-b)/2a which would need to = 0 for it to be symmetrical about the y axis. Therefore the numerator = 0 => b-6a = 0 => b=6a, answer (D). (Very real possibility of careless errors at any point in my workings).</p>

<p>U3NZIbk.jpg:
48: not sure.</p>

<p>bJw8Zh0.jpg:
10: (D) Three times, right? When x = 90, 180, 270, 2x = 180, 360, 540 so sin(2x) = 0.</p>

<p>NJbdXF8.jpg:
14: By definition, 1 radian is an angle that intercepts an arc equal to the radius of a circle, so if the arc = 12in, then the radius should be 12in as well so (D).</p>

<p>Vg23xAl.jpg:
18: No idea.</p>

<p>g6fJ04w.jpg:
I got (B) but it was a LOT of work. If you have a graphing calc, you could just graph the line and one circle at a time to see which combo touches. Here’s what I did (since I don’t have a graphing calc):</p>

<p>Sketched the line 3x-4y=10 (a.k.a.: y=(3/4)x-2.5). Sketched the line perpendicular to that and running through the origin (y=(-3/4)x). Calculated the co-ords of the point of intersection of those 2 lines. The line from the origin to that point is a radius of the circle.</p>

<p>…And after doing all thst I see you posted the answers as well so I shouldn’t have wasted time double checking. Oops.</p>

3

<p>I just went over 18 and 48 as lostint has pretty much answered the remaining questions.</p>

<ol>
<li>I am not sure if this is part of the SAT Math 2 syllabus. But it’s very similar to finding the radius of a circle. </li>
</ol>

<p>If you compare the given equation with the equation of a sphere ( x^2 + y^2 + z^ 2 + 2gx + 2fy + 2hz +c = 0 )
you’ll get g = 1, f = -2, h = 0, c = -10
The radius is sqrt(g^2 + f^2 +h^2 - c) = sqrt(1 + 4 + 10) = sqrt(15) = 3.87</p>

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

<p>Replace sin(theta) with y/r and cos(theta) with x/r.
You’ll get the equation in algebraic form to be x + y = 1
That means a straight line passing through (1,0) and (0,1). The x intercept is 1 and the y intercept is 1 as well.
The area between the x and y axes is simply the area of the triangle.
A = 1/2 ( 1 x 1) = 1/2 = 0.5</p>

4

<p>Seems like 21 hasn’t been done yet.</p>

<p>For 21, we want to find r such that 3x - 4y = 10 and x^2 + y^2 = r^2 has exactly one solution (x,y).</p>

<p>If you know how to compute distance from a point to a line, it turns out that the radius r is equal to the distance from 3x-4y = 10 to the origin, which is |-10|/sqrt(3^2 + (-4)^2) = 2, so the radius is 2.</p>

<p>Otherwise, it is possible to find r geometrically, noting that the x- and y-intercepts are A (10/3, 0) and B (0, -5/2). Let O = (0,0). The area of triangle AOB is (10/3)(5/2)/2 = 25/6. The hypotenuse AB has length sqrt((10/3)^2 + (-5/2)^2) = 25/6. Using area = base*height/2, the altitude from O to AB must have length 2, which is the radius of the circle centered at O. Therefore A is the answer.</p>

5

<p>@MITer94 - I did 21, but I forgot to label it. Also, you seem to have missed the last step (squaring the radius for the equation of the circle) so the answer is B.</p>

6

<p>@Iostint Whoops, yeah I should’ve checked, B is correct. The radius is 2, r^2 = 4. Thanks for pointing that.</p>

7

<p>@lostint @dialga2014 @MITer94 Thank you all so much for replying and even answering all the questions. Very greatly extremely appreciated.</p>