1.
Median is the middle one of and ordered set of numbers. There are many possible values for x, so let's just try the five given values.
(A) 2.5 - 5 = 5; 5 + 1 = 6; 3.5 - 8 = 7 5 < 6 < 7 OK
(B) 2.7 - 5 = 9; 7 + 1 = 8; 3.7 - 8 = 13 8 < 9 < 13 NOT OK
etc.
2.
Points (0,p) and (5,t) lie on the graph of x = y2 - 4. Substitute the given values for x and y; this tells you that
[1] ... 0 = p2 - 4, so p2 = 4, so p = -2 or +2
[2] ... 5 = t2 - 4, so t2 = 9, so t = -3 or +3
Line L therefore runs through points (0, +/- 2) and (5, +/- 3). To find the line with the steepest slope, we take the first point as low as possible and the second as high as possible: L goes from (0,-2) to (5,+3). The slope is
[3] ... [(+3) - (-2)] / [5 - 0] = 5 / 5 = 1.
3.
If the distance is D, then the time for both ways is
[morning] ... t1 = D / 45
[eventing] ... t2 = D / 30
We know that t1 + t2 = 1, so
D/45 + D/30 = 1
multiply by lcm of 30 and 45, that is, 90, we find
2D + 3D = 90
5D = 90
D = 18
If there are still steps you miss, mail me :)
2006-07-21 08:37:24
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answer #1
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answered by dutch_prof 4
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1)
A median is the middle number of a list of ordered numbers. For instance:
1 , 6 , 9.
6 is in the middle, so it is the median.
For this problem, plug in numbers and make lists and see which one fits the definition:
x = 5; 2x-5 , x+ 1 , 3x - 8 which is 5, 6 ,7. A is the answer
Try B just to see why its wrong:
x = 7; 2x-5 , x+ 1 , 3x - 8 which is 9, 8 ,13. A is the answer
Ordering the numbers, 8 ,9 ,13, 9 is the median, and 2x - 5 gave us 9, which contradicts the question.
2)
x = y^2 - 4
At (0,p) -----> 0 = y^2 - 4 ---> y = +/- 2, so p = +/-2
At (5,t) ------> 5 = y^2 - 4 ---> y = +/- 3, so t = +/-3
The possible points of intersection are:
(0,2) and (5,3); slope = (3 - 2)/(5 - 0) = (1/5) OR
(0,-2) and (5,3); slope = (3 - (-2))/(5 - 0) = (1) OR
(0,2) and (5,-3); slope = (-3 - 2)/(5 - 0) = (-1) OR
(0,-2) and (5,-3); slope = (-3 - (-2))/(5 - 0) = (-1/5)
The greatest slope is (1)
3)
Consider the time, distance and rate for each trip:
To work:
distance = d [m]
time = s [h]
rate = 45 [m/h] = d/s [m/h]
From work:
distance = d [m] (took same route, so the distance is the same)
time = t [h]
rate = 30 [m/h] = d/t [m/h]
They also tell you the total time is 1 hour
So, s + t = 1 [h]
In total we have:
45 = d/s
30 = d/t
s + t = 1
Divide second rate by the first rate:
(30/45) = (d/t) / (d/s)
(2/3) = (d/t) * (s/d) (reciprocal rule for dividing fractions)
2/3 = (s/t)
t = (3/2)s
Now substitute into s + t = 1
s + (3/2)s = 1
(5/2)s = 1
s = (2/5) hour = 24 minutes getting to work
t= (3/2)s = (3/2)*(2/5) = (3/5) hour = 36 minutes returning
Find "d" by using 45 = d/s, the rate for the "to work" trip.
45 [m/h] = d / (2/5) [h]
d = 45 [m/h] * (2/5) [h]
=18 m
2006-07-21 11:52:21
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answer #2
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answered by Anonymous
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On the first question if x + 1 is the median then it would be between the other two, so A is the only answer that could even make sense. Plug in your answers. If this is SAT that would be the way to go.
On the second question, plug in your two points to your equation
0 = p^2 -4 and 5 = t^2 -4 and solve for p and t.
p^2 = 4, so p is either 2 or -2
t^2 = 9 so t is either 3 or -3
now use the slope formula for the two points
and m = (t-p)/(5 -0)
use your options for t and p
put in 3 for t and -2 for p
m = (3- -3)/(5 - 0)
m = 5/5
m = 1
if you use the other values you get a smaller slope.
For the third problem, set 45t = 30(1-t)
solve for t
t = 2/5, in hours
distance equals rate times time
so 2/5 times 45 equals 18 miles.
2006-07-21 12:42:41
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answer #3
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answered by msmath 1
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[1] For (x + 1) to be the median, it must lie (inclusively) between the other two expressions.
2x - 5 ≤ x + 1 ≤ 3x - 8.
Consider each of the inequalities.
2x - 5 ≤ x + 1
x ≤ 6
x + 1 ≤ 3x - 8
-2x ≤ -9
x ≥ 4.5
Therefore 4.5 ≤ x ≤ 6. Of the possible values you're allowed for "x," 5 is the only one that works.
~~~~~ ~~~~~ ~~~~~ ~~~~~
[2] x = y² - 4
One point of intersection is (0, p), so
0 = p² - 4
p² = 4
p = ±2.
The other point of intersection is (5, t), so
5 = t² - 4
t² = 9
t = ±3.
The slope of line L (Δy / Δx) = (t - p) / (5 - 0).
The possible values are:
[3 - 2] / 5 = 1/5,
[3 - (-2)] / 5 = 1,
[-3 - 2] / 5 = -1, or
[-3 - (-2)] / 5 = -1/5.
The greatest of these values is 1.
~~~~~ ~~~~~ ~~~~~ ~~~~~
[3] Remember the rate × time = distance. This problem can be solved with two variables or with one. (I think two is easier.)
In the morning, she drives 45 mph (rate) for "m" hours (time). Her distance is (rate × time) 45m.
In the afternoon, she drives 30 mph for "a" hours. Her distance is 30a.
You know that her total time driving is 1 hour, so
m + a = 1.
You know both distances are the same, so
45m = 30a. Solve.
m + a = 1
a = 1 - m
45m = 30(a)
45m = 30(1 - m) [Substituting]
45m = 30 - 30m
75m = 30
m = 30 / 75 = 2 / 5 = 0.4
Remember that "m" is the number of hours she drove in the morning. The question is asking for the distance.
rate × time = distance
45(m) = D. [Substitute in 0.4, the value you solved for m.]
45(0.4) = D = 18.
The distance is 18 miles.
2006-07-21 09:04:38
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answer #4
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answered by Anonymous
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In part (3), we know that Esther drove to work in the morning 1.5 times faster than she did in the evening. Therefore she spent 1.5 times more times driving in the evening. 60:2.5=24, so she drove for 24 minutes in the morning and 36 minutes in the evening. Driving for 24 minutes at a speed of 45 miles per hour she would drive 45 * 24/60, or 18 miles.
2006-07-21 09:01:24
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answer #5
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answered by torturapatente 1
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it truly is a touch stupid to be making this evaluation. The SAT problem try for Math 2 is really a lot more effective not easy than the mathematics component to the SAT Reasoning try. in case you study for the mathematics 2 try and understand the fabric completely properly, then there is surely no rationalization why you should war with the mathematics component to the SAT, because it truly is going to be somewhat of cake compared to what the mathematics 2 exam throws at you.
2016-11-25 00:32:34
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answer #6
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answered by Anonymous
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1) 5
The numbers are
5, 5, 7
If it were any other number, then x+1 would not be the median.
2006-07-21 08:31:39
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answer #7
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answered by MsMath 7
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