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05-28-2006, 05:33 PM
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#1 | | New Member
Join Date: Jan 2006
Posts: 10
| Hard Math Problems- Need some help I know the answers but I don't know you can get that answer. If anyone can help me out with 1 or more problems, I would appreciate it.
These problems are from the blue book:
Pg 522, #17
1) In the xy-plane, line L passes through the origin and in perpendicular to the line 4x+y=k, where k is a constant. If the 2 llines intersect at the point
(t, t+1), what is the value of t?
a) -4/3
b) -4/5
c) 3/4
d) 5/4
e) 4/3
#20
2)When 15 is divided by the positive integer k, the remander is 3. For how many different vaules of k is this ture?
a)1
b)2
c)3
d)4
e)5
pg 534. #13
3) Let the function f be defined by f(x) = x+1. If 2f(p)=20, what is the value of f(3p)?
#15
4) A meauring cup contains 1/5 of a cup of orange juice. It is then filled to the 1 cup mark with a mixture that contains equal amounts of orange, grapefruit, and pineapple juices. What fraction of the final mixture is oranje juice?
Picture problems
pg 551, #16
pg 532, #8
Again, any help will be appreciated.
If anyone wants to know the answers I will post them. |
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05-28-2006, 06:12 PM
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#2 | | Member
Join Date: May 2006 Location: California
Posts: 538
| 1) slope of line L = 1/4 so equation of line L is y=x/4. plug (t, t+1) into y=x/4 and solve for t.
2) find values where the remainder equals 3, there are patterns to this. when 15 is divided by 12, the remainder is 3 so you find factors of 12 that cannot be divided into a number between 12 and 15. k=12, 4, 6 so C?
3) f(p)=10 so p= 9. f(3p)=28
4) 3/4 is filled w/ O, G, and P juice. so 1/4 is filled with each. there is already 1/5 so 1/4+1/5=9/20
I don't have the book so i can't do the rest. i'm not sure if these are the right answers, but that's how i would approach the problems. |
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05-28-2006, 06:33 PM
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#3 | | Junior Member
Join Date: May 2006
Posts: 33
| what are the answers, may be i can explain |
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05-28-2006, 08:04 PM
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#4 | | New Member
Join Date: Jan 2006
Posts: 10
| Thanks x3rose.
Here are the answers:
1) a) -4/3 , sorry but can you be more detailed when you say plug t, t+1 in
2) c) 3
3) 28
4) actually it's 7/15 or .466 (difficultly is medium)
9/20 = .45, I think this would be wrong
I'll post some pics of the other 2 problems tommorrow.
Thanks a lot.
Last edited by Nikesh; 05-28-2006 at 08:11 PM.
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05-28-2006, 08:16 PM
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#5 | | Member
Join Date: May 2006
Posts: 640
| I think he means plug the answer choices in for number one. |
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05-28-2006, 08:18 PM
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#6 | | Senior Member
Join Date: May 2006
Posts: 1,828
| Ok for number forget k
You get 2 equations
y = 4x
y = 1/4x
now plug in the points
T + 1 = 1/4T
-3/4T = 1
T = -4/3
For number 4. i see what she did wrong.
The first fraction is 1/5
There is 4/5 left and that is further split into 3 equal parts because there are 3 different juices. So you get 4/15 of each juice
So you get 1/5 + 4/15 = 7/15
Changed 
Last edited by Gyros321; 05-28-2006 at 08:33 PM.
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05-28-2006, 08:20 PM
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#7 | | Member
Join Date: May 2006 Location: California
Posts: 538
| oh yes gyros is right. sorry i can't subtract. yeah, i can do calculus but can't do basic math.. quite sad.
btw, i'm a girl. |
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05-28-2006, 08:38 PM
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#8 | | New Member
Join Date: Jan 2006
Posts: 10
| Thanks gyros321 and everyong for helping me. For #4, I misread the problem.
I'll be posting a couple of more problem up. I've been doing intense studying over the SAT. It hasn't let me slept in a month. It's so strange. How one test can decide everything. Oh well, there's always a retest.
I've been scoring around 690 on practice test on math. I have 5 more days till I take the SAT. |
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05-28-2006, 09:49 PM
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#9 | | Member
Join Date: Apr 2006
Posts: 563
| 551 #16
Since it asks for a rectangular region of 12L by 10L units wide, we know that the area will be 120L^2. We are given FIVE rectangles of dimensions L by W.
Since we're looking for an L^2 term, we seek for a way to find W in terms of L.
Remember that opposite sides of a rectangle are equal, so in the picture 2L = 3W. Solving for w gives us w=2L/3.
Be careful here, we need to find the number of rectangles of dimension L x W to cover an area of 120L^2. Read carefully to make sure you aren't solving for five rectangles instead of individual ones.
So we have w. L x W = L x 2L/3 = (2L^2)/3. Dividing the total area by the number of individual rectangles, we get the number needed. So the answer is E) 180
For #8, you could find the exact function, but notice that the outputs of function f are equal. Thus, f(2)=f(0), f(1) is of course = f(1), and f(3) = f(-1).
Remember outputs are y values and inputs are x values when dealing with coordinate geometry. Answer is C
Last edited by snipez90; 05-28-2006 at 09:51 PM.
Reason: ambiguous terms
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05-29-2006, 12:11 AM
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#10 | | Senior Member
Join Date: Aug 2004 Location: Student @ JHU (Baltimore, MD) / Houston, TX
Posts: 2,147
| You've gotten some good answers; this thread has a list of previously discussed questions that you might find useful. |
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05-29-2006, 10:39 AM
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#11 | | Member
Join Date: Mar 2006
Posts: 328
| For 532 #8
-The graph shows a parabola
-The only information you need here is in the last sentence(If f(b) = f(3), which of the following could be the value of b?)
-So basically, you're trying to find 2 values for x that when plugged into the function f(x) will equal the same y value.
-so when x = 3, y = 5(you can do this visually by tracing along the parabola)
-now all you have to do is find the other point in the graph that has the same y value and this point has an x value of -1(ans-c). The coordinates would be (-1, 5) |
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05-29-2006, 11:21 AM
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#12 | | New Member
Join Date: Jan 2006
Posts: 10
| Thanks for all the help.
I'll check that thread out tanman, before posting questions. |
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05-30-2006, 11:55 AM
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#13 | | New Member
Join Date: Jan 2006
Posts: 10
| I've check out the thread and I didn't find these problems.
I've been missing 6-7 total on math-which gives me a range of 660-720.
anyways, here are the problems:
Picture problems
pg 586, # 18
pg 599 #17
This was the answer from someone else:
"here's the thing. you'll have 11 line segments because besides V there are other 11 vertices, but V is conected with 3 vertices directly(meaning through edges). so your answer is 11-3=8.
hope you understand"
I don't get what the question is asking. I know that V is connected with 3 vertices directly but what does that have to do with, "How many of these segments will not lie on an edge of the figure." |
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05-30-2006, 02:00 PM
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#14 | | Junior Member
Join Date: Jul 2005
Posts: 237
| 586, #18:
We are told AB=BC and DE = EF=DF. That means the big triangle ABC is isosceles and the smaller triangle DEF is equilatreral. The angle measures of an equilateral triangle are all 60 degrees, so label them 60.
We are given ABC is 30 degrees, so label that, and label angles BAC and BCA 75 degrees. You know this because the triangle is isosceles, and 180-30 = 150/2 = 75 degrees.
We are also given angle BDE is 50 degrees, so label that.
The only other thing we need to know in order to calculate the measure of angle DFA is angle ADF - you know this because it is a supplementary angle (adds up to 180 degrees) with BDE (50 degrees) and FDE (60 degrees). If you don't see why this is, notice that they are on a straight line.
Therefore, angle ADF is 70 degrees (because 50+60+70=180), and angle DFA is 35 degrees because it is the third angle of the triangle with a 70 degree and 75 degree angle in it.
So the answer is B, 35 degrees.
599 #17:
In geometry, an "edge" is where two sides meet. For example, the four solid slanted vertical lines that connect the top base to the bottom base are all edges - as are the sides of the hexagonal top base.
Now that you know what an edge is, you can understand the question: it asks, how many segments can be drawn connecting point V to the other verticles in the figure, which do not lie on the edges of the figure?
Looking at the figure, you can see that there are only 3 straight line segments which would lie (completely) on edges of the figure. That would be, the line from V to the vertice directly under it, and from V to the two vertices directly left and right of it. All other lines connecting V to the other vertices (and you can try drawing these lines if it helps you visualize it) must travel over a side of the figure, or through empty space, in order to connect.
Notice the distinction between "side" and "edge."
Since there are 3 that lie on edges, and 11 different lines being drawn (one for each vertice) the answer is 11-3 = 8.
Hope that makes more sense. |
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05-30-2006, 04:42 PM
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#15 | | New Member
Join Date: Jan 2006
Posts: 10
| Thanks obsessedAndre for breaking it down. It makes the problem a lot easier to understand. |
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