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In the figure above, the smaller circles each have
radius 3. They are tangent to the larger circle at
points A and C, and are tangent to each other at
point B, which is the center of the larger circle.
What is the perimeter of the shaded region?

DrSteve

The diameter of a smaller circle is the radius of the larger circle. So the larger circle has radius 6. The circumference of the large circle is then 12pi, and the circumference of a small circle is 6pi. Now if we travel along the perimeter of the shaded region we trace out half the circumference of the large circle, then half the circuference of a small circle, then half the circuference of a small circle once more. So the answer is 12pi/2 +6pi/2 + 6pi/2 = 12pi.

jasonjackson789

@ Dr. Steve: thank you again!! Your explanation is freakin' awesome

cosmicomic

I'm having problems with these questions:

http://i.imgur.com/N3nB1.png
http://i.imgur.com/tIdno.png

Thanks in advance!

teteatete

16) (D) 3 and 1/2. Really doesn't matter which path you follow. Either make a staircase or go down then to the right.

17) (A) Six. Two left, two down; two down, two left; one down, two left, one down; one left, two down, one left; one left, two down, one left; one down, two left, one down.

18) (D) Because otherwise you cannot have an integer distance.

15) (p+n)(p-n) = 12 where (p+n) > (p-n). Right off the bat you know it can't be III, since that would mean p+n = 3. Now if p - n = 1, p + n = 12, so 2p = 13, p = 6.5, n = 5.5. That's fine. If p - n = 2, p + n = 6, so 2p = 8 and p = 4, so n = 2. That works too. The answer is (C) I and II.

It's always good to test things out for SAT math - better safe than sorry, right?

20) Ok, so if j, k, and n are consecutive, so if the units' digit of jn is 9, j must end in 9 too... e.g. 9, 19, 29... etc, and n will be 1, 11, 21, etc. k, then, will be something like 10, 20, 30, etc. so its units digit is 0. SWAG. The answer is (A)!

Europeen

I, too, am having trouble with a rather misleading problem.

<Any 2 points determine a line. If there are 6 points in a plane, no 3 of which lie on the same line, how many lines are determined by pairs of these 6 points?>

My answer was 30. This is because every point can form a line with one of the five other points, meaning the possibilities are 6 x 5 (assuming one point can be part of multiple lines).
I feel the difficulty lies in deciphering the prompt, is my interpretation of "how many lines are 'determined'" correct? What does the prompt mean, exactly?

Any help would be much appreciated!
-euro

rspence

@Europeen you have to divide by two, because you are counting each line twice. For example, you counted P6P5, and P5P6, which are the same line. The answer is 6*5/2 = 15 (or 6C2).

tranman123456

A list consists of 1000 consecutive even integers. What is the difference between the greatest number in the list and the least number in the list?


Answer is................................................ .................................................. ..............1998.
I got 2000. Because I had 0 as my first even integer. Then 2000 as my largest. I got 2000 as my largest even integer because I noticed that 20 was the 10th consecutive integer and 30 was the 15th consecutive integer so to get the 1000th even integer I multiplied 1000 by 2. Where did I go wrong? Thanks in advance!

arooj1a2b3c

There is probably a more simple way to do the above problem, but I used the arithmetic sequence formula: a(n) = a1 + (n-1)d. (d = number of increase, a1 = first number, n = term number

a(1) = a1 + (1-1)(2)
a(1) = a1 (first even integer and thus the least number in the list)

a(n) = a1 + (1000 - 1)(2) (1000 represents the final and greatest term of the list)
a(n) = a1 + (999)(2)
a(n) = a1 + 1998

a(n) - a(1) = difference of greatest number and least number
a1 + 1998 - a1
= 1998

Europeen

@tranman: You know where you went wrong? If you take 0 as your first and 2000 as your last value you create a range of 1001 integers not 1000.
Imagine it this way: there are 2000 consecutive integeres from 1 to 2000. If you take half of them it makes 1000 even (or odd, whatever the prompt says) integers. Your smallest value would be 2 (or 1 if it's about odd integers). Zero is not included! If you take zero as your smallest value your highest one is 1998, not 2000. 1998 - 0 = 1998 so it's correct again.

Europeen

First, sorry for the double post. Unfortunately, I am unable to edit posts on my cell phone.

I have stumbled across an SAT problem that perplexes me profoundly. Here it goes:

<If a, b, c and f are four nonzero numbers, then all of the following proportions are equivalent except...>

A) a/f = b/c
B) f/c = b/a
C) c/a = f/b
D) a/c = b/f
E) af/bc = 1/1

There is no restriction on these numbers except they are unequal to zero. Heck, they could be everything! I could arbitrarily assign numbers to the equations that every single one of them works. What are we looking for here?

MuffinMan17

^^Is the answer E?

HelloParis

1. The median of a list of 99 consecutive integers is 60. What is the greatest integer in the list?

A: 109. What's the quickest & most easiest way to get this answer?

MuffinMan17

@HelloParis, since there are 99 integers and 60 is in the middle, you can conclude that 60 is the 50th integer and that there are 49 integers before it and 49 integers after it (49 + 49 = 98, and with 60 in the middle, that makes 99 integers). So 60 + 49 = 109

HelloParis

@MuffinMan17 Thank-you! That makes much more sense now!

Can anyone explain this question to me?

1. Let the function f be defined by f(x) = 5x for all numbers x. Which of the following is equivalent to f(p +r)?

A: 5p + 5r (Why is not 5p + r?)