Test 2 Section 5 - #6 (page 464)

The key here is recognizing that slope = rise/run (or, in this question, y/x).  According to this question the slope of that ramp is 7/16.  We can actually BACKSOLVE this one, by basically taking our answers and plugging them into rise/run (or, like we said, y/x) to see which one will give us that same fraction.  We know our rise (or y) is 3.5 feet.  With backsolve, always try (C) first:

(C) 3.5/20 = 7/40.  Nope.  Way too long on bottom.  Go smaller.
(B) 3.5/12.5 = 7/25.  Still too long on bottom, but getting warmer...
(A) 3.5/8 = 7/16.  BINGO.

Test 2 Section 5 - #14 (page 468)

Tough one (grid-ins usually are). The solution to this one, though, can be found by PLUGGING IN (with a little BACKSOLVE flavor added in for good measure)! Just start by trying any number for x. How about x = 3? Just to be clear, we're looking for f(3) to be LESS THAN 3 (that's what they say in the question: f(a) < a).

f(x) = |3x - 17|

f(3) = |3(3) - 17|

f(3) = |9-17|

f(3) = |-8|

f(3) = 8

Ok, so that didn't work; 8 is not less than 3. But here's where that BACKSOLVE flavor comes in: now we have a better idea of how the pieces fit together. We want x to get a little bigger, so that 3x gets closer to 17.  That way, f(x) will be small.  Why not try x = 5 next?

f(5) = |3(5) - 17|

f(5) = |15-17|

f(5) = |-2|

f(5) = 2

GREAT! 2 < 5, so x = 5 works! (A few other numbers will work too: anything between 4.25 and 8.5...though you'd probably be best to stick with integers and just try 5, 6, or 7.)

Test 2 Section 5 - #16 (page 468)

For questions like this, it's best to just start listing the numbers that are "tri-factorable" (by listing numbers that are made by multiplying 3 consecutive positive integers, starting with the lowest positive integers we can). It won't take us long:

1 x 2 x 3 = 6
2 x 3 x 4 = 24
3 x 4 x 5 = 60
4 x 5 x 6 = 120
5 x 6 x 7 = 210
6 x 7 x 8 = 336
7 x 8 x 9 = 504
8 x 9 x 10 = 720
9 x 10 x 11 = 990
10x11x12 = 1320  (Oops, too high!)

So how many were under 1000?   9 of them.

Test 2 Section 5 - #15 (page 468)

This one has hints of PLUGiIN and BACKSOLVE, even though it's a grid-in! We don't need any of that stuff about how many candies are in the jar...that's just extraneous information. All we need to know is where Ari's starting from (3 red, 4 green) and how many more he takes. We want him to have more red pieces than green pieces by the end, but just barely. Here's how I set it up:

RED	GREEN	Comments
3	4	This is our starting point.
+7	+6	Just plugging in 2 numbers that are close together and add up to 13.
=10	=10	OOPS!  That's not enough red...Ari has the same amount of each color now.

Now it's kinda clear where to go: add one more red (and one less green):

RED	GREEN	Comments
3	4	This is our starting point (again).
+8	+5	Plug in new numbers.
=11	=9	That'll do it!  Any less red and he'd have the same number of each (as we saw above).

Test 2 Section 5 - #18 (page 468)

Just fill in what you know and work it out:

You know the side of each square is k. So that means the AREA of each square will be k². Since there are ten squares total, that means the ENTIRE area will be 10k². So a=10k².

Now get the perimeter: just count the number of sides that are bolded. There are 16 sides to the figure, and since each side equals k, that's 16k. So p=16k.

Since the question says p=a, that means 16k=10k². Divide both sides by k to cancel it out, and you get 16=10k. Then divide by 10 on both sides and you get 16/10=k, which reduces down to 8/5=k.

Test 2 Section 5 - #17 (page 468)

First write out the expressions for each of the two carriers:

Carrier A is $1.00 for the first 20 minutes and $0.07 for every minute after 20, so that expression for the total cost would be: 1+0.07(t-20). (The "t-20" is how you write the phrase "numbers of minutes past 20.")

Carrier B is simply $0.06 per minute, so that expression would be: 0.06t.

The question asks for when the costs would be the same, so set the expressions equal to each other: 1+0.07(t-20) = 0.06t. Distribute to get 1+0.07t-1.4 = 0.06t, and combine like terms to get: 0.01t = 0.4. Solve that to get t = 40.

Test 2 Section 5 - #7 (page 465)

If you have a graphing calculator, this one is probably easiest to do using Plug-In. First, plug in anything you want for a... let's go with a=3. Now graph that equation: y=3x²+2. Next graph the SECOND equation (y=x²+2) and compare the two graphs. You'll see that the second graph is wider than the first, so the answer is (B).