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Old 02-28-2005, 02:55 PM   #1
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math ?s from new sat book

#7 on 473

#18 on 412

#14 and 15 on 427

#16 on 475

#17 on 522

#8 on 596

#18 on 657

#15 on 684

Any help would be greatly appreciated.
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Old 02-28-2005, 09:44 PM   #2
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bump............
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Old 02-28-2005, 10:16 PM   #3
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i would help u if i had the book
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Old 03-01-2005, 12:53 AM   #4
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post the questions...
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Old 03-01-2005, 02:37 PM   #5
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Same here... put some efforts to type them out please

I just came in, saw the numbers, and went out.
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Old 03-01-2005, 04:32 PM   #6
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16, p. 475:

h(x) = 14 + (x^2 / 4)
h(2m) = 14 + ((2m)^2 / 4)
h(2m) = 14 + (4m^2 / 4) <---you can cancel the 4s here or get a common denominator and then factor out ... the latter is more work and the former works just fine)

h(2m) = 14 + m^2
Set this = to 9m...

14 + m^2 = 9m
Factor away!
0 = m^2 - 9m + 14
0 = (m - 2)(m - 7)
m = 2, 7 <---CB will accept either one

15, p. 427

Split PQ into 2 segments (since we know a quadratic curve with its axis running along the y-axis will be symmetric about the y-axis... lol)... So we have two segments of length 3. This gives us our x value(s) for the y = x^2 function. f(3) = (3)^2 = 9. Since the other function intersects at this same point, we can set that function = to 9...
9 = a - (3)^2
9 = a - 9
18 = a

7, p. 473 has something to do with breaking the numbers down; some sort of trick that I forget how to do...
8, p. 596 has to do with trying numbers in the expression... Someone can probably provide a better explanation... The answer is 3^3 * 4^2 which = 432.

14, p. 427 ... I just multiplied both sides of the equation by (a - b)^(1/2). Then I squared both sides of the equation, and I got a^2 - b^2 = 1.

17, p. 522... I have no idea
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Old 03-01-2005, 06:48 PM   #7
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"7, p. 473 has something to do with breaking the numbers down; some sort of trick that I forget how to do..."

18 sqrt 18 is the same as 18 srqrt 2*9 or 18*3 sqrt 2

The answer is thus 54 * @ or 108.
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Old 03-01-2005, 06:51 PM   #8
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18 on 412

Calculate (2*30*45)/(30+45) => 2700/75 equals 36.
Now I have the average speed of 36 mph for one hour of travel. Since the distance IS the same both ways (duh), the distance is 36/2 or 18 miles.

PS The formula is 2 * speed1 * speed2 / speed 1 + speed 2. I could explain the formula, but it is not necessary since the formula is easy to remember and ALWAYS works. Astute students will also notice that the formula presents a variance: multiply both speeds and divide by the straight average of the speeds given. For people who use their head before jumping to the calculator, this may allo for a quicker number manipulation. For students interested in super shorcuts, you should also know that if this type of problem appears on a MC, the answer is ALWAYS slighly below the straight average. The straight average is ALWAYS the trick question.
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Old 03-01-2005, 07:04 PM   #9
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18 on 657

1. To solve it you need to remember that (a-b)^2 = a^2 -2ab + b^2

2. Then, look at what was given by the ETS maniacs... You need to find a way to reduce one of the functions. The value of t=0 seems interesting because of the zero. Plug in the function for a height of 6:

6 = c-(d-0)^2 or c = 6 + (d-0)^2 or C = 6 + d^2 <= remember this one

3. Now that we have C, we can plug it in the function for 106:

106 = (6+d^2) - (d-10)^2 ... play around a bit and you'll find that d = 10.

4. Now that you have c AND d, you can simply write the function for t = 1. The answer is indeed 70.

It takes a bit of manipulation. As usual, make sure not to miss any clues. In this case, the t=0 was an important clue and not an afterthought.
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Old 03-01-2005, 07:13 PM   #10
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15 on 684

To find the answer, you'll have to construct or visualize two right triangles that include the midpoints.

The first one will create a hypothenuse that joins the corner below B and the point A. It is quite easy to find the value since you have two sides of 2 and 1. Now, you simple create another right triangle with a hypo that joins A and B. You have the value for the right sides as the first calculated value (it is sqrt 5) and 1.

Leave the values in root form (based on a quick look at the proposed solutions). The answer is sqrt 6.
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Old 03-01-2005, 07:14 PM   #11
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Post number 7 should be

The answer is thus 54 * 2 or 108 and not The answer is thus 54 * @ or 108.
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Old 03-01-2005, 08:39 PM   #12
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Thanks for the help guys, I sure am rusty.

17 on 522

In xy plane line t passes through the origin and is perpendicular to the line 4x + y= k, where k is constant. If the two lines intersect at the point (t,t+1) what is the value of t?
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Old 03-02-2005, 07:40 AM   #13
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If lines are perpendicular, the product of their slopes is -1.

Line_2 can be rewritten as y = k - 4x, so it has a slope of -4, and Line_1 therefore has a slope of -1/-4 = 0.25
You can write Line_1 as y = n + 0.25 x, where n=0 (since the line passes through the origin). Plug in x=t, y=t+1 into this equation:

t+1 = 0 + 0.25t, from which t = (-1) / 0.75 = -4/3 .
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Old 04-27-2008, 01:01 PM   #14
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proble 16, page 684

I dont get how to solve this problem?
How does the x corresponde to the a?
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Old 04-28-2008, 02:11 AM   #15
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Look up 684 / 16 / in http://talk.collegeconfidential.com/...ns-3rd-ed.html
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