Explain these couple of Math questions?

Test Preparation
1

<p><a href="http://i62.tinypic.com/xp4kxv.jpg"&gt;http://i62.tinypic.com/xp4kxv.jpg&lt;/a>
The answer is A, the farthest I can get up to finding out the angles of the mini-triangles and I don't know what to do from there</p>

<p><a href="http://i60.tinypic.com/2rgc87m.jpg"&gt;http://i60.tinypic.com/2rgc87m.jpg&lt;/a>
Answers are E and B, I don't know where to begin for the first one except plugging in the x and y values which led me nowhere, and the second one all I could do was discover the sides of the rhombus which are 6.</p>

<p><a href="http://i61.tinypic.com/2qnshkz.png"&gt;http://i61.tinypic.com/2qnshkz.png&lt;/a>
The answer is C, now there doesn't seem to be any number crunching involved, so what's the rule/principle/concept that I don't know here?</p>

2

<h1>1. Triangle AFG is an isosceles triangle therefore the base angles are congruent. If one angle is 22 then the other two must be 79 each. The parallel lines will make the corresponding angles congruent, therefore x=79</h1>

3

<h1>1: You should have angle AGF = 22 deg, GAF and GFA = 79 deg (GA congruent to GF, so angles are congruent–isosceles triangle). From there, you can get angle BFG = 101 deg (180-79) and angle FGC = 158 deg (180-22). Then, you can get angle GCB = 22 deg because BG and FE are parallel, and the same line cuts both of them, and angles AGF and GCB are in the same relative positions as they relate to the transversal (the line intersecting with each respective horizontal line). Thus, x = 360 - 101 - 158 - 22 = 79 deg = choice A.</h1>

<p>(The concept you would use to get GCB = 22 deg is the same as Joe’s way of getting x. His way’s a lot easier…but I didn’t see it before I posted. :slight_smile: )</p>

4

<h1>2. Parabolas opening up have leading coefficient or “a” positive or a>0</h1>

<p>can’t read the shaded region value in second question</p>

5

<p>Last one, not sure best way to explain this one, it is just the North-south diameter, which would be two radii.</p>

6

<h1>2:</h1>

<p>19: If the leading coefficient (in this problem, “a”–the coefficient of the term with x^2) of a quadratic is greater than 1, the graph will have gone fewer than one x-unit away from the x-coordiate (on either side) of the vertex one y-unit away from the y-coordinate of the vertex. </p>

<p>If the leading coefficient (the coefficient of the term with x^2) of a quadratic is less than 1 but greater than 0, the graph will have gone greater than one x-unit away from the x-coordiate (on either side) of the vertex one y-unit away from the y-coordinate of the vertex.</p>

<p>If the leading coefficient is negative, the same is true for all the above relative to -1, the graph just opens downwards. </p>

<p>If the leading coefficient is 0, there’s no x^2 term, and the graph becomes a straight line. Just something helpful to know. But if you didn’t know all that?</p>

<p>You essentially have three points: (-2, -1) along with two that you can reasonably assume: (-1.5, 0) and (-2.5, 0). You can plug those values in for x and y in a system of three equations with three unknowns.</p>

<p>20: Can you slide the figure over any more so that the perimeter of the shaded area in the text is visible? 'Twould be appreciated…but I’ll assume it’s 24, because you said the sides of the rhombus were 6. Here’s what I would do: I would draw line segment XY. You now have two triangles: each with two radii of two (seemingly congruent) circles, and XY. Well, golly good gracious, those are equilateral triangles: XY, per the figure, is a radius too! </p>

<p>Thus, angles X and Y of the rhombus are each 120 deg. Circumference of one circle = 2<em>pi</em>r = 2<em>pi</em>6 (remember, one side of the rhombus, which is a radius, is 6). 360-120 deg = 240 deg. 240 / 360 = 2/3. Thus, one shaded arc is 2/3<em>12</em>pi = 8pi. The two shaded arcs, together, then, are 8pi + 8pi = 16 pi = choice B.</p>

7

<h1>3: Joe said it pretty well. Ignore the plane itself, and focus instead on the circular cross-section of the sphere made by the plane (the circle on the plane with center P). You can think of that as the ground, and all possible radii as a stake. How many different ways can you put in the stake so that the stake is only straight up and down–not driven into the ground at any other angle than straight in–with one end on the ground (an end only–nothing in the middle of the stake)? Two: one with the bottom on the ground and the top sticking straight up, and one with the top on the ground and the bottom pointing straight down.</h1>

<p>Hopefully that helps more than confuses you. Just trying to think of other ways of looking at it.</p>

8

<p>7:57PM · in SAT and ACT Tests & Test Preparation
<a href=“http://i62.■■■■■■■.com/xp4kxv.jpg”>http://i62.■■■■■■■.com/xp4kxv.jpg&lt;/a&gt;
answer is A. in TriangleAGF, x = <AFG = (180-22)/2 = 79.</p>

9

<p>Apologies, for the 3rd question, the perimeter is 24 </p>