Question

if possible find the slope of the line interpret the slope in terms of rise and run. A) the slope of the line is __________. B) the graph falls _______ units for each 4 units of run. Graph plots are (0,5) and (4,0)

Answer 1

Okay, when we go from (0,5) to (4,0) we've gone 4 units along the x-axis, aka "the run". We from 5 to 0 along the y-axis, or -5 units, aka "the rise". Slope is the ratio of rise / run.

Answer 2

so I got the slope as -1.25 and that the graph falls 4 units of run am I correct?

Answer 3

One of those is correct, one is not.

Answer 4

the graph falls 1.25 units for each 4 units of run

Answer 5

Look at the problem statement again more carefully. the graph falls _____ units for each 4 units of run.

Answer 6

Slope is the number of units risen or fallen per unit of run.

Answer 7

Also, you go from (0,5) to (4,0) — the run is 4. what was the change in y over that distance?

Answer 8

1.25

Answer 9

does 5 - 0 = 1.25?

Answer 10

no

Answer 11

it would be 5

Answer 12

right. well, you go from y = 5 to y = 0. what is the rise?

Answer 13

yeah, if you meant -5, that's correct. It goes from 5 to 0 in the space of 4 units on the x-axis. that means the slope is (0-5)/4 = -1.25, and the graph falls 5 units for every 4 units of run.

Answer 14

so am I correct on the slope at least?

Answer 15

you've got the slope correct. perhaps a picture would help reinforce the second half. make yourself a little graph showing the line through (0,5) and (4,0) and convince yourself that the rise/fall is -5 when we go from x = 0 to x = 4

Answer 16

ok i have graph already made out but I am still confused on how to find the rise per run

Answer 17

Just count over some convenient number of spots along the x-axis (I usually go however many it takes to get the line to exactly cross an intersection or tick mark on both ends of my range). That's the run. Now count the change in y from where it was at the beginning of the run to the end of the run. The ratio of rise to run is the slope. If you know the slope, and you know the run, you can compute the rise. Or if you know the slope, and you know the rise, you can compute the run.

Answer 18

Ok I think I got it now

Answer 19

If you look at your graph, and you have to move 2 units to the right for the line to go up 1 unit, that means you've got a rise of 1 and a run of 2 and a slope of 1/2. With that slope, if you have a run of 8, you'll have a rise of 1/2 * 8 = 4.

Answer 20

got it now

Answer 21

If you had a run of 1, with a slope of 1/2, the rise would be 1/2*1 = 1/2.

Answer 22

thanks

Answer 23

you're welcome!

Answer 24

how are you with exponents?

Answer 25

fabulous :-)

Answer 26

when working with negative exponents how do I do that

Answer 27

They work just like positive exponents. Just remember that \[x^{-n} = \frac{1}{x^n}\]

Answer 28

my problem is (9m)\[^{-2}\]

Answer 29

So, if you have a negative exponent, and it's in the denominator, get rid of the minus sign in the exponent, and put it in the numerator. If you have a negative exponent in the numerator, get rid of the minus sign in the exponent and put it in the denominator.

Answer 30

\[(9m)^{-2}\]

Answer 31

\((9m)^{-2}\)
?

Answer 32

sorry still getting use to the equation functions on here

Answer 33

just fabulous

Answer 34

the answer that I was coming up with was .012346m

Answer 35

Okay, that's easy. Using my "template" from before, x = 9m, and n = -2.
\[x^{-n} = \frac{1}{x^n} \rightarrow (9m)^{-2} = \frac{1}{(9m)^2}\]

Answer 36

but that just doesnt seem right to me

Answer 37

Good. it's not :-)

Answer 38

81m?

Answer 39

So \[(9m)^{-2} = \frac{1}{(9m)^2} = \frac{1}{(9)^2m^2} = \frac{1}{81m^2}\]

Answer 40

oh ok I see now

Answer 41

You might also write that as \[\frac{1}{81m^2} = \frac{m^{-2}}{81} \approx 0.0123457m^{-2}\]

Answer 42

Notice that I took m^2 out of the denominator so that I didn't have a fraction to write, but it is still really a fraction...

Answer 43

that might be how I was coming up with the .012347m just wasnt putting the squared

Answer 44

yeah I seen that, it made it look a lot easier without it, i understood it better

Answer 45

Well, no, it's not just that - note that I have a negative exponent, not a positive...
it really does represent \[0.0123457 * \frac{1}{m^2}\]

Answer 46

The difference between what you first wrote and what I wrote is a factor of m^3...

Answer 47

ok

Answer 48

That's a substantial difference, so you want to avoid making that mistake, or one like it.

Answer 49

I will make note of that, thank you so much, you have been teaching me more tonight than my instructor has in the last 3 months

Answer 50

Like many things, it just takes a bit of attention to detail. Work methodically, and check your work.

Answer 51

agreed