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08-17-2005, 01:44 PM
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#1 | | Member
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| SAT Math Questions
Page 412 Blue book
#17
In the xy-coordinate plane, the graph of x = ysquared -4 intersects line l at (0,p) and (5,t). What is the greatest possible value of the slope of l?
answer: 1
#18
Esther drove to work in the morning at an average speed of 45 mph. She returned home in the evening along the same route and averaged 30 mph. If Esther spent a total of one hour commuting to and from work, how many miles did Esther drive to work int he morning?
answer: 18
If anyone could explain how to get these answers thatd be great. | 
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08-17-2005, 01:57 PM
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#2 | | Member
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| bump anyone know how to do them | |
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08-17-2005, 02:10 PM
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#3 | | Senior Member
Join Date: Apr 2005 Location: Canada; Penn 2012
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| #18
Let x represent the number of hours she drove.
45x = 30(1-x)
45x = 30 - 30x
75x = 30
x = 0.4
Now put substitue 0.4 for x
45(0.4) = 18.
30(1-0.4) = 18 | |
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08-17-2005, 02:12 PM
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#4 | | Senior Member
Join Date: Jul 2005 Location: Bloomington IN
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| 17) The graph of x=y^2-4 is a sideways hyperbola. To find the "p" of for x=0, just plug it in:
0=y^2-4
4=y^2, y=2 or -2
Likewise, 5=y^2-4, 9=y^2, y=3 or -3
You know the hyperbola connects between (0,2) and (5,3) and between (0,-2) and (5,-3). Looking at a graph, the steepest line is between (0,-2) and (5,3).
The slope of that is: (-2-3)/(0-5)=-5/-5=1 | |
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08-17-2005, 02:12 PM
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#5 | | Senior Member
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| For #18
The formula for her average speed is (2*Speed1*Speed2)/(Speed1+Speed2)
So it is (2*45*30)/(45+30) = 36. Since the total time is one hour, the total distance travelled is also 36 miles, and te distance for half of the trip is 36/2 = 18 miles. | |
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08-17-2005, 02:15 PM
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#6 | | Senior Member
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| Find p and t. x=y^2-4, so 0=p^2-4. 5=t^2-4. 4=p^2, 9=t^2. p=+/-2 and t=+/-3
Slope equals (change in y)/(change in x); use the greatest possible difference in the change in y (which is equal to t-p) which would be (3-(-2)) to get 5/5=1
d=rt
d=(r1)(t1)=(r2)(d2)
45(t1)=30(t2)
t1+t2=1
45(1-t2)=30(t2)
45-45(t2)=30(t2)
75(t2)=45
t2=.6
d=(r2)(t2)=30*.6=18mi
Or you could use the harmonic mean...faster but this method is more general. | |
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08-17-2005, 02:23 PM
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#7 | | Member
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| Hyperbola? Shouldn't it be a parabola? And what about (5-0)/(3-2)? That comes out to 5. | |
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08-17-2005, 02:28 PM
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#8 | | Senior Member
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| Yea, it is a parabola (sideways). | |
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08-17-2005, 02:29 PM
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#9 | | Senior Member
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| Whoops. It is parabola. I've been out of school too long 
That would be [5-0]/[3-(-2)]=(5-0)/(3+2)=5/5=1 | |
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08-17-2005, 02:30 PM
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#10 | | Member
Join Date: Dec 2004
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| A hyperbola is a parabola sideways? I thought it was the points where the distance of all was 2c from the locus or something like that?
Why can't you do the points (0,2) and (5,3)? | |
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08-17-2005, 02:34 PM
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#11 | | Senior Member
Join Date: Jul 2005 Location: Bloomington IN
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| Na, hyperbola is like x^2-y^2. It's like two big parabolic-like functions in opposite directions. There's a gap where points are "undefined" because one arm goes in one direction to infinity, the other goes in the other direction. I think there was something to do with a locus too, but that was back in AlgebraII. | |
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08-17-2005, 02:35 PM
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#12 | | Senior Member
Join Date: Jul 2005 Location: Bloomington IN
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| If you do the points (0,2) and (5,3) you get (2-3)/(0-5)=-1/-5=1/5. The question is for the greatest slope, and 1 is greater than 1/5. | |
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08-17-2005, 02:37 PM
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#13 | | Member
Join Date: Dec 2004
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| Ohhh yeah. Thanks. | |
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