<p>Oh… so you didn’t know what unhindered interaction meant.</p>
<p>Kinda lol</p>
<p>If a and b are positive integers and [(a^1/2)(b^1/3)]^6 = 432, what is the value of ab?</p>
<p>I thought I could do (ab)^5/6 x 6 = (ab)5 = 432 then take the fifth root of 432, but that doesn’t work, so what did I do wrong and how do I solve it 
…
Here is a question that I flatout didn’t understand what they were asking: in a rectangular coordinate system, the center of a circle has coordinate 5,12 and the circle touches the x axis at one point only. What is the radius of the circle? </p>
<p>Is a rectangular coordinate plane the same as a 3D rectangle? If it is then I’m also guessing the single point that it touches is in the center, making the radius 12.</p>
<p>
</p>
<p>[(a^1/2)(b^1/3)]^6 = 432; distribute the sixth power
(a^3)(b^2) = 432
Prime factorization of 432 = 2<em>2</em>2<em>2</em>3<em>3</em>3 or (4^2)(3^3)
a = 3; b = 4
ab = 12</p>
<p>
</p>
<p>A rectangular coordinate system is the same as the x-y plane. You are correct. The answer is 12.</p>
<p>Nevermind I see the x axis is just tangent to the bottom of the circle! Derp <-</p>
<p>Page 733, #14</p>
<p>14 I see that the final answer is the product of the two inequalities, but I don’t know why I multiply them.</p>
<p>
</p>
<p>0 ≤ x ≤ 8 and -1 ≤ y ≤ 3</p>
<p>Okay… Think about it. The greatest possible value of xy occurs when x is 8 and y is 3. The least possible value occurs when x is 8 and y is -1. Therefore, -8 ≤ xy ≤ 24 is the correct answer.</p>
<p>yeah i feel like it would be harder because they call it a hard question</p>
<p>anyways on this test i got a 47/54 raw score so that’s a 660-720 =) The fact that i got that means its prolly closer to 660 though ;)</p>
<p>Page 764 #10</p>
<p>I see C as correct because the point that the author makes is that the women were counted for purposes of representation, but could not vote. Being free is certainly an issue, but in the context of that line, does not support the answer. </p>
<p>By the way how is this an easy question!? I got every one right except that one and that’s marked “Easy” :@</p>
<p>
</p>
<p>C is deliberately wrong. Read the answer choice and lines 11-12.
Answer choice reads: suggest that women could be appointed as representatives but could not vote.</p>
<p>Appointed as representatives means “elected to office”.</p>
<p>The quotation was used to emphasize the irony in this predicament where women were “free” but without political rights.</p>
<p>Side comment: I think we’re working through the BB at a similar pace.</p>
<p>Lol ^^ Thanks, I didn’t notice the appointed. </p>
<p>p 773 #18</p>
<h1>15 i also dont know how to solve, but CB loves that type of problem</h1>
<p>18.
h(t) = c - (d-4t)²;
6 = c - (d-4(0))²;
6 = c - d²;</p>
<p>h(t) = c - (d-4t)²;
106 = c - (d-4(2.5))²; Expand
106 = c - (d² - 20d + 100);
106 = c - d² + 20d - 100;
206 = c - d² + 20d; Plug in c - d² = 6
206 = 6 + 20d
200 = 20d
d = 10
c - d² = 6
c - 10² = 6
c = 106</p>
<p>Now that the two constants are figured out, the new equation is h(t) = 106 - (10-4t)²</p>
<p>h(1) = 106 - (10-4(1))²;
h(1) = 106 - 36 = 70</p>
<p>15.</p>
<p>3 Plumbers: 1 experienced and 2 trainees.
There are 4 experienced and 4 trainees.</p>
<p>How many possibilities are there for 1 experienced? 4 possibilities
How many possibilities are there for 2 trainees? <a href="43">b</a>/2! trainees**
4 * (43)/2! = 24 possibilities</p>
<p>You have to really understand counting to do well on SAT Math!
Either you know how to utilize combinations and permutations or you can think logically.</p>
<p>This counting is still messing me up. Where did the three come from?</p>
<p>There are 4 trainees to choose from.
First trainee: 4 possible choices
Second trainee: 3 possible choices</p>
<p>4<em>3 = 24 possible choices
Let’s say the four trainees are A, B, C and D
However, in this process, you double counted AB and BA. Therefore, you must divide 4</em>3 by 2. = 6</p>
<p>6 possibilities for trainees and 4 possibilities for experienced.</p>
<p>6 * 4 = 24</p>
<p>Why am I not double counting BC and CB or anything like that</p>
<p>I just don’t get these… it’s really weird because I have taken 6.5 of these practice tests and my score isn’t improving at all. I might just resign to being bad at taking the SAts</p>
<p>
</p>
<p>Oops. Sorry. I don’t think I was clear in the other post.
You are double counting AB, BC or BC, CB or CD, DC or AC, CA and etc… You are absolutely right.
In total, you double counted 6 of them.</p>
<p>Don’t be discouraged. Maybe your studying method isn’t right. For every 20 minutes you spend solving problems, you should spend 40 minutes reviewing your incorrect answers thoroughly. (why did you get them wrong? what was the mistake in your thought process? or anything). If you don’t review your incorrect answers and just go on, you’re basically taking one step forward and one step backward. Einstein defined insane as doing the same thing over and over again and expecting different results. Try changing your studying method :).</p>
<p>In short, this is what I’m saying:
You have to first understand counting problems before you go on to solve another one.</p>
<p>Try solving this problem and explain your thought process:</p>
<p>There are 5 points in a circle A, B, C, D, and E. A line is created when two of these points are connected. How many different lines can be created? (Note that AB and BA are not the same.</p>
<p>Another one:</p>
<p>Jerry is at an ice cream shop. First, he has to choose a cup size. Then, he must choose two different toppings. There are four different cup sizes and 6 different toppings. How many possible combinations are there?</p>
<p>Last one: (variation of the second one)</p>
<p>Jerry is at an ice cream shop. First, he has to choose a cup size. Then, he must choose three different toppings. There are four different cup sizes and 6 different toppings. How many possible combinations are there?</p>
<p>If you can solve these three problems, you can be sure that you have conquered the SAT counting problems.</p>
<p>Oh. If you don’t know how to approach the problems, ask me. I’ll do one for you.</p>
<h1>1 is 20</h1>
<h1>2 is 60</h1>
<h1>3 is 80</h1>
<p>I think I got these right, although I used a calculator method with probability. That isn’t how the CB wants people to do it, though :p</p>
<p>Going to do them without a calculator now and attempt to explain how I got them.</p>
<h1>1:</h1>
<p>First points: 5 possible choices
Second point: 4 possible choices</p>
<p>Don’t divide by 2 because AB and BA are not the same. </p>
<h1>2:</h1>
<p>Cup Size: Four possible sizes (4)
Toppings: Six choices total and 2 toppings ([6 x 5]/2 = 15)
Divided by two because I would have double counted some combinations. </p>
<p>15 x 4 = 60 possible outcomes</p>
<h1>3:</h1>
<p>I’m stuck on this one. ([6 x 5 x 4]/3?) x 4 </p>
<p>Same number of cup choices, so I must need to divide by 6, not 3. Why do I divide by 6 instead?</p>
<p>3 is a little overboard. You probably won’t see this one on SAT.</p>
<p>You are correct in questioning why you must divide by 6. Think about it. For every 3 topping choices, you are double counting 6 same combinations. (ABC, ACB, BAC, BCA, CAB, CBA)</p>
<p>If it helps, think if it as doing a counting within the counting.</p>
<p>You divide 120 by 6. Note that 6 is equivalent to 3! (three factorials).</p>
<p>With understanding of counting, you can write your own formula.
n!/(r!)(n-r)!
6!/(3!)(6-3!) = (6<em>5</em>4)/(3<em>2</em>1) = 20 possibilities.
Simply you just divide by the (number of choices factorial).</p>