Blue Book, pg 361, help

heartonsleeve · 06-30-2007, 10:29 PM

Hey, can someone help me out? I can't quite figure out #12, 13, or 14 from the blue book, pg 339 and 340. Thanks in advance. (and sorry for the mistake in the thread title!!)

ny10804 · 06-30-2007, 10:53 PM

What is this blue book everyone talks about?

heartonsleeve · 06-30-2007, 10:56 PM

It's the Official SAT Study Guide by College Board.

DukeofFunk · 07-01-2007, 02:42 PM

Dunno if you need help, but as far as I can tell for 12 you have to do the problem with the rule that two sides of a triangle are always greater than the third side in mind. 3, 4, and 5 wouldn't work because they're all less than or equal to 7, but 6 would. Therefore, the answer is one possible triangle.

For 13, I can't figure out any quick way to do it besides actually solving for x (-1, -5) and y (5, -7) and then adding all the possible numbers and knowing that 10 doesn't exist. There's probably a quicker way to do this, but I'm just too stupid.

For 14, since a multiplied by (b-a) is 0, then a must be 0. It wouldn't matter whats in the parentheses, because whatever it's multiplied by makes it 0, and since a is the number on the outside, it must be 0. Since a is greater than b, and a is 0, b must be less than 0. If b is less than 0, and a is 0, then a-b will be > than 0 because the two negatives would cancel out. Like, if b was -3, it would be 0-(-3), or 0+3, which would make it greater than 0. So all three statements are true

Did that help? Did you still need help? Am i too late? Is this confusing for you? Sorry. I hope I helped.

heartonsleeve · 07-01-2007, 04:07 PM

Thank you!!! Yes, I did still need help.

I got 12 and 14 but I still don't understand 13. ^^;;
Could you explain that again?

DukeofFunk · 07-01-2007, 04:47 PM

Ok, well to solve the problem the long way, solve each equation individually. Get the abs value sign by itself, so

|x+3| = 2
|y+1| = 6

Since they're both within absolute value signs, you have to set equations for both positive and negative. So it'd be like

x+3 = 2 AND x+3=-2

ditto with the y equation. For x, that'll get you -1 and -5. For y, you'll get 5 and -7. Then since it's asking for all the possible x+y's, you add the different combinations together (-1 + 5, -5 + 5, -7 + -1, and -7 +-5) and take the absolute value of the number. So it'd be 4, 0, 8, and 12. You can't come up with 10, so that's the answer.

There's probably a simpler way to do this, but this is how I figured it out.

heartonsleeve · 07-01-2007, 04:58 PM

Thanks!! I really appreciate it

This is a really silly question, because I know I should know how to do this, but can you solve #7 on pg 361 for me? A similar problem came up in one of the next practice sets, and I couldn't figure out either.

Thanks a million!

DukeofFunk · 07-01-2007, 05:19 PM

The first set of points gives you the y intercept, making the equation y=1/4x + 1/2. Plug 4 from the other set into the equation, so 4(1/4) + 1/2 = 1 1/2.

heartonsleeve · 07-01-2007, 05:27 PM

Thank you!!! Again, I really, really appreciate it.

xFactorx · 01-21-2009, 05:53 PM

wouldn't x= -1 and y= 5 (not -5) ?

hmhanh · 01-23-2009, 06:15 AM

For #7 on pg 361, there's probably a faster way:
since u know the slope of a line is:
(the change in y between any 2 points on the line)/ (the change in x between the same 2 points on that line)
Hence:
(y-0.5)/ (4-0)=1/4
--> y= 3/2

Can anyone solve page 340 /No.15 (the parabola one) for me! Thks in advance ^

hmhanh · 01-29-2009, 04:52 AM

Quote:

Can anyone solve page 340 /No.15 (the parabola one) for me! Thks in advance ^

No one? ...

gcf101 · 01-29-2009, 10:46 AM

I've never seen a question on inverse functions on the SAT I. In 412/17, for example, you don't need to know the graph or properties of x = y^2 - 4.

Anyway, reflect the given graph in the line y=x; you'll get a parabola y=ax^2 - 2, where "a" is some coefficient
Swap x and y and drop a:
x = y^2 - 2.
It's (A).