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In the figure above, lines k, m and p have slopes r, s and t respectiv

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In the figure above, lines k, m and p have slopes r, s and t respectiv  [#permalink]

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New post 15 Dec 2017, 01:19
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A
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C
D
E

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In the figure above, lines k, m and p have slopes r, s and t respectively. Which of the following is a correct ordering of these slopes?

(A) r < s < t
(B) r < t < s
(C) s < r < t
(D) s < t < r
(E) t < s < r

Attachment:
2017-12-15_1300_002.png
2017-12-15_1300_002.png [ 8.2 KiB | Viewed 981 times ]

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In the figure above, lines k, m and p have slopes r, s and t respectiv  [#permalink]

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New post Updated on: 16 Dec 2017, 09:24
Bunuel wrote:
Image
In the figure above, lines k, m and p have slopes r, s and t respectively. Which of the following is a correct ordering of these slopes?

(A) r < s < t
(B) r < t < s
(C) s < r < t
(D) s < t < r
(E) t < s < r

Attachment:
2017-12-15_1300_002.png

We need to rank the lines' slopes in order of steepness.
A slope is defined as "bigger," or "steeper," by its absolute value.

Line k is horizontal, slope = 0
The least steep slope is k's, which is \(r\)
\(r\) should be the first variable in the answers.
Eliminate answers C, D, and E

A or B?
Which is greater, \(s\), line m's slope, or \(t\), line p's slope?

This diagram initially frustrated me.
With tired eyes, I couldn't tell which was steeper, line m or line p.
Tired brain, too. I forgot to use the basics.

The slope between two points on a line is the change in y (up or down) divided by the change in x (left to right):
\(\frac{rise}{run}= \frac{y_1-y_2}{x_1-x_2}\)

A good place to determine the coordinates of two points: a line's x- and y-intercepts.

On the diagram, it IS clear that line p's intercepts are farther from the origin than line m's intercepts.*
Estimate.

Line m has a y-intercept at about (0,1) and an x-intercept at (3,0)
Line p's y-intercept is higher, at about (0,2). Its x-intercept is (-3.5,0)

Slope of line m:
\(\frac{y_1-y_2}{x_1-x_2}=\frac{0-1}{3-0}=-\frac{1}{3} \approx{-.33}\)

Slope of line p:
\(\frac{y_1-y_2}{x_1-x_2}=\frac{0-2}{(-3.5)-0}=\frac{-2}{-3.5}\approx{.57}\)

The sign of the slope does not determine steepness.
(A negative slope is not smaller than a positive slope because it has a negative sign.)

Steepness is ranked by the absolute value of the slope.
|.57| > |-.33|

Hence
\(s < t\), and \(r < s < t\)

Answer

*Both x- and y-intercepts of p are farther from the origin than m's. Careful. That might mean the slope's ratio is smaller. Slope is change in y divided by change in x.

Edited content to reflect change in method. Answer unchanged.

Originally posted by generis on 16 Dec 2017, 00:19.
Last edited by generis on 16 Dec 2017, 09:24, edited 1 time in total.
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In the figure above, lines k, m and p have slopes r, s and t respectiv  [#permalink]

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New post 16 Dec 2017, 00:27
What's the answer?
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Re: In the figure above, lines k, m and p have slopes r, s and t respectiv  [#permalink]

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New post 16 Dec 2017, 00:30
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In the figure above, lines k, m and p have slopes r, s and t respectiv  [#permalink]

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New post Updated on: 17 Dec 2017, 03:49
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Bunuel wrote:
Image
In the figure above, lines k, m and p have slopes r, s and t respectively. Which of the following is a correct ordering of these slopes?

(A) r < s < t
(B) r < t < s
(C) s < r < t
(D) s < t < r
(E) t < s < r

Attachment:
The attachment 2017-12-15_1300_002.png is no longer available


Given: k = r, m = s, p = t.

To find the slope of the line: \(\frac{(y2-y1)}{(x2-x1)}\)

Line k seems to be a horizontal line, which means the y coordinate is static (always the same value), so the slope r = 0.
Line m is going downwards, so the slope = s is negative.
Line p is going upwards, so the slope = t is positive.

So, s < r < t = negative < 0 < positive (assuming that k is a horizontal line).

(C) is the answer.
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Originally posted by exc4libur on 16 Dec 2017, 09:23.
Last edited by exc4libur on 17 Dec 2017, 03:49, edited 1 time in total.
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Re: In the figure above, lines k, m and p have slopes r, s and t respectiv  [#permalink]

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New post 16 Dec 2017, 19:44
A line that swings upward has positive slope, and a line that swings downward has a negative slope. The slope of a horizontal line has a slope of 0. It's obvious that s<r<t. The answer is C.
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In the figure above, lines k, m and p have slopes r, s and t respectiv  [#permalink]

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New post 16 Dec 2017, 22:45
exc4libur and Paulli1982 :

I have learned that the sign of a slope just tells us which direction it goes, not whether it is steeper or flatter.
I have learned that the slope of a line tells us the rate of change of y relative to corresponding change in x.
So it is hard for me to understand the reasons you answered the way you did.
As far as I can tell, you rest your case on the number line alone. Is that correct?

As I said, I learned that the slope of a line tells us the rate of change of y relative to x.
That seems to imply that whether its slope is positive or negative, a line that is close to vertical has a very steep slope.
It also seems to imply that whether its slope is positive or negative, a line that is close to horizontal has a very gradual slope.

If the slope is 4, y increases rapidly compared to x.
If the slope is -4, y decreases rapidly compared to x.

Given those two mathematical relationships, it would seem that the magnitude of the slope, not its sign, determines steepness.
True, the sign tells you whether it goes "right and up", or "right and down."
But I cannot see how a positive or negative sign tells us how quickly y increases or decreases in relation to x.

Stated again: The slope of a linear function is a rate of change.

A horizontal line, with a slope of zero, does not change. At all. Ever.

Think of a skateboarder.
First she is on a flat sidewalk. Zero slope.
Then she climbs up the left side of a hill facing that sidewalk.
She skateboards down the right side.

Are you willing to argue that she deals with a greater slope while on a flat sidewalk than while going down the hill?
If so, would you please explain why?

Or suppose there is a train.
It goes up a perfectly symmetrical mountain, which looks like this: /\
The train makes it to the top. At the top, it loses its brakes.
Now it is a runaway train. It is hurtling down the other side of the mountain.
Because, and only because, it hurtles downward and to the left, (and has a slope of, say -5), would you say its slope while hurtling without brakes is LESS than the slope on flat ground?
If so, would you please explain why?

I respect disagreement. That said, Paulli1982 I do not think there is any need for words such as "obvious."
If it were so obvious, I would not be taking a different position.

Please explain to me how and why a horizontal line with no change (y is always the same) has a greater slope than a line that looks like this: \

Please explain how to account for the rate of change in your scenario.

Please explain how the sign of the number denoting of a rate of change in a linear function tells me how steep my hike will be, both up the mountain and down?

I think magnitude of change, or absolute value of the number that describes rate of change, determines greater or smaller slope, steeper or less steep slope.
I think a slope of -4 is greater than a slope of \(\frac{1}{2}\). Why? It has a greater rate of change.
Greater rate of change? Steeper slope.
Smaller rate of change? Flatter slope.

I am very curious to know how you are measuring steepness. And why. Maybe I am missing something.

If there is some way to order slopes about which I do not know, I would appreciate knowing.

Thanks in advance!
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Re: In the figure above, lines k, m and p have slopes r, s and t respectiv  [#permalink]

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New post 17 Dec 2017, 03:48
genxer123 wrote:


genxer123

Hey, you do have a point, if the ordering of the slopes is determined by its absolute value then yes, I agree with you.
But, since the answer could be (A) or (B), then the question would have to give us more information.

Since it doesn't give us any more information, I think its not asking for the ordering of the absolute value of the slopes, its just asking for the ordering of the values of "m" in the slope form equation: \(y=mx+b\).
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Re: In the figure above, lines k, m and p have slopes r, s and t respectiv &nbs [#permalink] 17 Dec 2017, 03:48
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