<p>@ everyone
Critical reading sucked!!How was it??</p>
<p>The height increased in increments of 4, GenericMath. I, too, thought it was increasing in multiples of 2…BUT…the figure was drawn to scale, and it seemed evident the added line segment in each height was definitely not double the previous distance (which was 4).</p>
<p>My answer anyway is 80.</p>
<p>@Jassica; yes, no/18’s answer was 120.</p>
<p>Let’s see other opinions on this.</p>
<p>I made answer for last one was 120 aswell.</p>
<p>for the tirangle one yes me 2 did it 80 :)</p>
<p>^Sir</p>
<p>Yes but figure drawn to scale he didn’t present any information about the other subjects.
So we use the information that we know acctually exists which is apparent
Since each of the smaller triangle had a base of 5,so we 2x had base of 10,so
5 + 5 = 10;
5 + 4 = 9,so based on this information 5 + 4 != 10.</p>
<p><a href=“http://img214.imageshack.us/img214/1803/30388817.png[/url]”>http://img214.imageshack.us/img214/1803/30388817.png</a></p>
<p>did anyone get this answer 36pi</p>
<p>i did it my the same way …</p>
<p>^ Which question are you talking about?</p>
<p>Since people will be asking about number 20, here goes.</p>
<p>What is the units digit of 7^1600?
The key here is to look for a certain pattern, and evidently there is.</p>
<p>7^1 = 7
7^2 = 49
7^3 = A very long number that ends with 4.
7^4 = A very long number that ends with 1.</p>
<p>And the pattern keeps repeating itself. Different numbers, but same units digit. 7, then 9, then 4, then 1. Therefore, now to know which unit digit 7^1600 is at, we divide 1600 by 4. The remainder is 0. Therefore, the units digit of 7^1600 is 1.</p>
<p>I see makes sense @ Sir.</p>
<p>Yes I solved that question same way.</p>
<p>So far, I’m at -0. Let’s hope it stays this way.</p>
<p>What about you, GenericMath?</p>
<p>Never mind its 4 not 5 numbers I didn’t notice it properly.</p>
<p>[7^1,7^2,7^3,7^4,7^5,7^6,7^7,7^8,7^9,7^10</a> - Wolfram|Alpha](<a href=“7^1,7^2,7^3,7^4,7^5,7^6,7^7,7^8,7^9,7^10]7^1,7^2,7^3,7^4,7^5,7^6,7^7,7^8,7^9,7^10 - Wolfram|Alpha”>7^1,7^2,7^3,7^4,7^5,7^6,7^7,7^8,7^9,7^10 - Wolfram|Alpha)</p>
<p>No, there was question about finding the area of the shaded area(a segment i think).radius 12 anyone remember??</p>
<p>options were </p>
<p>24 pi
36 pi</p>
<p>sumthing like that</p>
<p>So far I am -2</p>
<p>For the parabola question:
Which one of these lines cuts the equation x^2 -5bx + 6 at two points?
First we need to understand what this equation means. It means that the vertex of the parabola is at (5,0). It also means that it cuts the x-axis at 6. It also means that this parabola has a lowest point. Now, let’s move onto the given choices. Which one is going to cut this figure (after you draw it) at two points?</p>
<p>I) Y = 6. Yes. It will. Since the y-intercept of this parabola is 6, then indeed, a line that is parallel to the x-axis at Y = 6 will cut the parabola at two points.
II) X = 2. No. It is a vertical line. Since when do vertical lines cut parabolas in two points?
III) Y = X + 1. If you draw it, you’ll see that it’s impossible for this line to cut the given parabola at two points.</p>
<p>The only available choice that makes sense is (I). Anyone confirm?</p>
<p>@imagallagher:
60/360 x Pi(12)^2 = 24Pi</p>
<p>Oh I remember that question I made it was equilateral triangle.</p>
<p>Since all of them has the same side,6,6,6 I think</p>
<p>So all angles inside are 60.</p>
<p>60/360 = 1/6 * 144 = 24PI I think would like hear Sir opinions on this.</p>
<p>That answer is correct, GenericMath, refer to my post above yours.</p>
<p>@Sir Yes I made it (I) only.</p>
<p>Alright. Here’s another one.
n/20
f(x)= 4ax^2 + bx + 4c</p>
<p>a+c = 15</p>
<p>Find the value of f(1) - f(-1)?</p>
<p>4a(1)^2 + b(1) + 4c = 4a + b + 4c
4a(-1)^2 + b(-1) + 4c = 4a - b + 4c</p>
<p>Add both equations, resulting equation: 8a + 8c. Take 8 common factor. 8(a+c) = 8(15) = 120.</p>
<p>Confirmed?</p>
<p>@Sir </p>
<p>You know which writing was actually the experimental one?
I am sure I made 100 % in the one about school boy entering university.
Do you know if that was the experimental one?</p>