<p>You multiply 12 by 11 to find the number of groups of 2 out of 12 people. For the first person, there are 12 possibilities, and for the second person, there are 11 possibilities. (You can’t have a person be the first and second choices.) Therefore you must multiply 12 by 11.
Next, you have to divide by 2. From what I did before, each pair of students can be chosen in two different ways. For example, student A and student B can be chosen as A then B, or B then A. To get rid of these extra pairs, we divide our answer from the first step by 2.</p>
<p>Oh, it’s to get rid of the extra pairs. Thank you! By the way, you’re probably going to be seeing a lot of me in this thread. :)</p>
<p>No problem. It’s fun to review math, especially since I’m pretty much only focusing on CR right now.</p>
<p>Awesome! Ok, I have another couple of questions for you.
I found this quick formula in Xiggi’s thread:</p>
<p>[2<em>Speed1</em>Speed2] / [speed1 + Speed2]</p>
<p>Would I only use this in problems where a person goes traveling the same distances but in different rates and times? Would the formula still work if I replaced “speed” with time? </p>
<p>ALSO</p>
<p>I saw another formula in the same thread:</p>
<p>sum of an arithmetic sequence = (first term + last term)/2 * (number of terms)</p>
<p>I tried using it to find the sum of the positive odd integers less than 100, and my calculator gave me 1, which is definitely not the sum. Can someone show me an example using this formula in action?</p>
<p>Thanks!</p>
<p>I’m not too sure about Xiggi’s formula, since I don’t have much experience on it. I’m more of a traditional d=rt person. Why don’t you ask the great Xiggi himself? 
I saw him around on CC just the other day.</p>
<p>
</p>
<p>Here is an easy example you can easily verify: Find the sum of the positive odd numbers less than 6. (1, 3, 5, and the sum is 9)
First term + last term: (1 + 5) = 6
Divide by two: 6 / 2 = 3
Multiply by number of terms: 3 * 3 = 9
It works!</p>
<p>Now lets do it on your example: Find the sum of the positive odd integers less than 100.
First term + last term: (1 + 99) = 100
Divide by two: 100 / 2 = 50
Find the number of terms: 100 / 2 = 50
Multiply by number of terms: 50 * 50 = 2500
Well, looks about right, and its certainly not 1. :)</p>
<p>Oh, what I did was multiply 2 and the number of terms, and put that whole answer under the denominator. I read the formula incorrectly haha. Also, yeah I guess I’ll PM Xiggi, I didn’t think he was still on! Most people on CC seem like they are here for a year and then just disappear. :p</p>
<p>Thank you!</p>
<p>Can someone kindly help me with a few problems? 
Referring to parabolas, what is the minimum value of f(x)=(x-1)(x-5)</p>
<p>For what positive value of k does the graph of y=x^2+kx+9 intersect the x-axis exactly once?</p>
<p>If f(x)=x^2-7, which of the following is the reflection of f(x) about the y-axis?
a) y=f(x)
b) y= -f(x)
c) y= lf(x)l (absolute value)
d) y=f(-x)+7
e) y= f(x+7)</p>
<p>If x^2-ax+7=x^2+x-b for all values of x and if a and b are constants, what is the value of a-b?</p>
<p>ty!</p>
<p>For the parabola you just foil it and get x^2-6x +5. the formula for the vertex is b/2a . so (-6/((2)(1)))=3. then u plug in 3 into the equation and get -4. You could also graph it.</p>
<p>for the second one you have to have a perfect square equation. (x+3)^2 and when u foil you get x^2 +6x +9. 6 is the answer. alternatively u could try graphing with various numbers to see that.</p>
<p>for the third one it is D. just graph it. </p>
<p>For the last one u just match the numbers. -a=1 because they are connected to the x’s and 7=-b since they are the constants. so you get a-b=-8.</p>
<p>a fitness center purchased a number of exercise machines 4 costing 1700 8 costing 1300 x costinf 1200 if the median is equal to1300 and x is an odd integer what is thre greatest value for x ?</p>
<ol>
<li>Because 4+8=12, x must be odd in order for the median to equal 1300. Now, can ignore the 4 units costing 1700 for now. Calling the 8 machines a b c d e f g h, we maximize x by making a to be the median, 1300. So to cancel bcdefgh when determining the median, we must have 7 machines that cost 1200 however me must not forget the 4 machines that cost 1700 so our answer is 4+7= 11 as desired.</li>
</ol>
<p>Hello, first post I had toruble with these questions while taking some oractice tests. My math skills are a bit rusty at the moment. Any help would be really appreciated </p>
<p>1) The length and width of a rectangle have integer
values. If the area of the rectangle is 75, what is one
possible value for the perimeter of the rectangle?</p>
<p>2)
A D 1 4 7
B E 2 5 8
C F 3 6 9</p>
<p>A snack machine has buttons arranged as shown above.
If a selection is made by choosing a letter followed by
a one-digit number, what is the greatest number of
different selections that could be made?</p>
<p>1) So the length and width can’t be fractions. Area is 75. Assume two values for length and width. We’ll go with 15 and 5, in which 15 x 5 = 75.
Therefore, one possible value of the perimeter is 2(15+5) = 40.</p>
<p>2) 6 x 9 = 54 different selections. Am I correct? If so, I solved it like so: How many different choices do you have for a letter? Six. A, B, C, D, E, and F. How many different choices do you have for a number? Nine. 1, 2, 3, 4, 5, 6, 7, 8, 9. Now, to get the total number of different possibles letter number combinations, you multiply both sums (6x9) which gives a product of 12. This question is related to combinations/permutations, and largely depends on your knowledge of their rules. Review this math section in the BB or any prep. book of your choice.</p>
<p>Yes, your answers are correct for both questions Thank you so much for explaining. I have reviewed permutations and combinations from the BB before but I’m still having trouble with the topic. Is Gruber’s going to be helpful for the slightly trickier math questions?</p>
<p>The only thing that will improve your proficiency in permutations/combinations, regardless of the prep. book, is practice. Review the rules, memorize everything you need to memorize, and practice practice practice.</p>
<p>Good luck.</p>
<ol>
<li><p>P to P’ (3,-1) by a glide reflection for the translation (x,y)
to (x-3, y) and reflection across the line y=2. What are the coordinates of P?
(HINT: Work backwards to find the pre-image coordinates)</p></li>
<li><p>What is the reflection image of (a,b) across the line y=-6?
A. (a-6, b) B. (a, b-6) C. (-12-a, b) D. (a, -12-b)</p></li>
</ol>
<p>In post #39:
That’s what happened to me when I revisited the following question (post #35):
For the number of votes for candidates I and II we’ll use respectively variables CI and CII.<br>
Their average A is obviously 50% of their sum.
Let’s mark on the number line points CII, A, and CI (A is of course smack in the middle between CII and CI).
The distance between CI and CII is 28,000, which is 1% of 2.8 million, so
CII is .5% to the left of A, and
CI is .5% to the right of A - that makes it 50.5%. The end.</p>
<p>Okay, this is from Gruber’s Complete SAT Math Workbook, page 242, question #6. I’ll write the question anyway in case you don’t have the book.</p>
<p>Mr. Smith drove from point A to point B in 2 hours. He drove back from point B to point A in 3 hours. If the distance from point A to point B is 100 miles, what is Mr. Smith’s average speed for the entire trip?</p>
<p>The first time I did this, I got the answer right, which is 40 miles. But then, several weeks later I went back to do it again and completely forgot how to solve it. No matter what method I try, I always end up getting either 20 or 24. Cam someone show me how to to this step by step?</p>
<p>
</p>
<p>For this problem, you’re going to want to use the harmonic mean formula. Don’t let the name intimidate you, it’s just a formula used to find the average speed when someone travels the same distance at two different rates. Here’s the formula itself:</p>
<p>Average speed = (2(Speed1)(Speed2))/(Speed1+Speed2)</p>
<p>Well first we have to figure out speeds 1 and 2. If from A to B is Speed1, then the speed is 100/2 = 50 mph. If from B to A is Speed2, then 100/3 = 33.3 mph. Now we can plug in these values into our formula:</p>
<p>Average speed = (2(50)(33.3))/(50+33.3) = 3333.3/83.3 = 40 mph.</p>
<p>How often do average speed questions show up on the actual SATs? I mean, statistically?</p>
<p>^^ Since they gave you time and distance for both legs of the journey, you don’t need the harmonic mean. You traveled a total of 200 miles (100 each way) in a total of 5 hours… 200/5 = 40 mph</p>