nystudent23

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.

nystudent23

bump anyone know how to do them

CDN_dancer

#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

tkm256

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

Flipsta_G

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.

towerpumpkin

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.

il bandito

Hyperbola? Shouldn't it be a parabola? And what about (5-0)/(3-2)? That comes out to 5.

Flipsta_G

Yea, it is a parabola (sideways).

tkm256

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

il bandito

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)?

tkm256

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.

tkm256

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.

il bandito

Ohhh yeah. Thanks.