Tutor.com Linear Equations Session

Feb. 18, 2014

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Session Transcript - Math - Algebra, 1/27/2014 10:08PM - Tutor.comSession Date: 1/27/2014 10:08PM
Length: 117.8 minute(s)
Subject: Math - Algebra


System Message
[00:00:00] *** Please note: All sessions are recorded for quality control. ***

Guest (Customer)
[00:00:00] The table below shows the federal minimum wage in different years. year: 1960|1970|1980|1990|100| minimum wage: 1.00|1.60|3.10|3.80|5.15| a. Find the rate of change for each ten-year time period. Show your work. b. During which time period did the minimum wage increase the fastest? Explain what the rate of change for this time period means.

Kathryn C (Tutor)
[00:00:19] Hello and welcome

Guest (Customer)
[00:00:40] The last year is supposed to say 2000, sorry.
[00:00:48] hello

Kathryn C (Tutor)
[00:01:16] Give me a minute to look thru

Guest (Customer)
[00:02:34] ok

Kathryn C (Tutor)
[00:03:37] Just copying down.
[00:04:19] Are you familiar with slope?

Guest (Customer)
[00:04:36] Yes

Kathryn C (Tutor)
[00:04:57] That is what you are finding here.

Guest (Customer)
[00:05:18] slope and rate of change are the same thing?

Kathryn C (Tutor)
[00:05:26] yes they are

Guest (Customer)
[00:05:35] okay

Kathryn C (Tutor)
[00:06:25] So your first 2 points would be (1960, 1.00) and (1970, 1.60)

Guest (Customer)
[00:06:56] okay so I would do the y2 - y1 over x2 - x1?

Kathryn C (Tutor)
[00:07:02] yes

Guest (Customer)
[00:07:28] I'm doing it on my paper; I'll tell you what I get

Kathryn C (Tutor)
[00:07:34] ok

Guest (Customer)
[00:08:33] I got .60 over 10

Kathryn C (Tutor)
[00:08:44] Simplify

Guest (Customer)
[00:08:56] .10 over 1, so .10

Kathryn C (Tutor)
[00:09:17] divide .60 by 10

Guest (Customer)
[00:09:31] .6 is the slope

Kathryn C (Tutor)
[00:10:05] .60 divided by 10 is .06, isn't it?

Guest (Customer)
[00:10:37] yes sorry
[00:11:11] so the next step is y-y1 = m(x-x1) ?

Kathryn C (Tutor)
[00:11:29] No you don't need to use that.
[00:12:32] When you found the slope, you found the rate of change ($.06 increase per year) for the first ten year period.

Guest (Customer)
[00:13:19] so I keep doing that until I've found the rate of change for each time period?

Kathryn C (Tutor)
[00:13:27] yes
[00:14:04] 60-70 which you just did, 70-80, 80-90, 90-2000

Guest (Customer)
[00:14:32] okay

Kathryn C (Tutor)
[00:14:53] Do you want me to check when you finish each one?

Guest (Customer)
[00:15:24] yes please

Kathryn C (Tutor)
[00:15:28] ok

Guest (Customer)
[00:16:39] 1970 - 1980 is $0.15?

Kathryn C (Tutor)
[00:16:46] yes

Guest (Customer)
[00:16:53] ok

Kathryn C (Tutor)
[00:19:08] I'm working on this...

Guest (Customer)
[00:19:32] 1980-1990 is $0.07?

Kathryn C (Tutor)
[00:19:38] yes

Guest (Customer)
[00:21:45] 1990-2000 is $0.27?

Kathryn C (Tutor)
[00:22:23] Did you take 5.15-3.80?

Guest (Customer)
[00:22:34] yes

Kathryn C (Tutor)
[00:23:19] It should be 1.35, shouldn't it?

Guest (Customer)
[00:23:53] yes, and then simplified it is 0.27

Kathryn C (Tutor)
[00:24:21] Are you dividing 1.35 by 10?

Guest (Customer)
[00:24:25] by 5

Kathryn C (Tutor)
[00:24:46] From 1990-2000 is 10 years.

Guest (Customer)
[00:25:19] so you don't simplify?

Kathryn C (Tutor)
[00:26:02] Did you divide both numerator and denominator by 2?

Guest (Customer)
[00:26:52] I get what I did wrong. I simplified by 5 because it went into both 1.35 and 10,m but it's ten years so I don't do that and leave it as is

Kathryn C (Tutor)
[00:27:26] You can just divide 1.35 by 10

Guest (Customer)
[00:27:52] and then you would get 0.135

Kathryn C (Tutor)
[00:28:41] Yes and you could have reduced by 5 first. It would still work. .27/2

Guest (Customer)
[00:29:34] so the answer is $0.135

Kathryn C (Tutor)
[00:30:01] yes ( is 13 1/2 cents)
[00:30:35] So you found the rate of change for each time period.

Guest (Customer)
[00:31:35] okay and part two the answer is 1990-2000

Kathryn C (Tutor)
[00:32:06] What was the increase for 70-80?

Guest (Customer)
[00:32:33] 15 cents, oops.

Kathryn C (Tutor)
[00:32:42] That's ok.

Guest (Customer)
[00:33:30] Thank you! I have two more questions but if I need to connect to a different tutot I can

Kathryn C (Tutor)
[00:33:46] I'll help you.
[00:34:51] Did you answer the last part of your first problem?

Guest (Customer)
[00:35:06] yes.ok. the next one has a small graph and wants me to find the slope of the line for the first part. I
[00:35:29] can do that but part b is what I will need help on

Kathryn C (Tutor)
[00:35:35] ok

Guest (Customer)
[00:37:40] I got 1 over -2 as the slope

Kathryn C (Tutor)
[00:37:48] ok

Guest (Customer)
[00:39:17] part b says: write an equation in slope-intercept form for a line that is perpendicular to the line in part a and has the same y-intercept as the function in part a.

Kathryn C (Tutor)
[00:40:14] What is the y-intercept for your given line?

Guest (Customer)
[00:40:41] (0, 3)

Kathryn C (Tutor)
[00:41:22] What do you know about the slopes of 2 parallel lines?

Guest (Customer)
[00:41:47] they are equal

Kathryn C (Tutor)
[00:42:04] I'm sorry.
[00:42:12] Perpendicular lines.

Guest (Customer)
[00:42:39] the product of them is -1 and they are opposites (reciprocal)

Kathryn C (Tutor)
[00:43:20] Yes. First slope = -1/2, second slope would be ?

Guest (Customer)
[00:43:33] 2/1

Kathryn C (Tutor)
[00:43:50] And the y-intercept is 3.

Guest (Customer)
[00:43:55] yes

Kathryn C (Tutor)
[00:44:07] Can you write the equation for the second line now?

Guest (Customer)
[00:45:07] it would be in y=mx+b right?

Kathryn C (Tutor)
[00:45:12] yes

Guest (Customer)
[00:45:29] so is it y = -1/2x + 3?

Kathryn C (Tutor)
[00:45:43] That is the first line.

Guest (Customer)
[00:46:17] y = 2/1x + 3?

Kathryn C (Tutor)
[00:46:38] yes and you can write 2/1 as 2

Guest (Customer)
[00:48:05] the last one has a part a, b, c, and d. I actually feel like I've learned something thank you so much

Kathryn C (Tutor)
[00:48:31] You are very welcome. I am enjoying working with you.

Guest (Customer)
[00:49:06] :)

Kathryn C (Tutor)
[00:51:08] I'm working on this...

Guest (Customer)
[00:52:33] the last one says: there is a linear relationship between the wind speed at a given temperature and what that temp "feels like". A higher wind speed will make the temperature feel colder. The table below shows what an unknown temp (t) "feels like" at different wind speeds
[00:53:23] Wind Speed (mi/h): 5|10|15
[00:54:07] Feels Like (F): 36|34|32
[00:55:05] a. write an equation in slope-intercept form relating the wind speeds to the unknown temperature.

Kathryn C (Tutor)
[00:56:32] So your temperature depends on the wind speed.

Guest (Customer)
[00:57:29] yes

Kathryn C (Tutor)
[00:58:31] temperature would be x and the wind speeds would be y

Guest (Customer)
[00:58:44] ok

Kathryn C (Tutor)
[00:59:24] The temperature is dependent and that is always your "x" value.
[00:59:47] So find your slope.

Guest (Customer)
[01:00:06] for each one like we did with the years?

Kathryn C (Tutor)
[01:02:08] I'm working on this...
[01:02:03] yes

Guest (Customer)
[01:03:09] so for (36, 5) (34, 10) I got the slope to be 5/-2

Kathryn C (Tutor)
[01:04:13] yes

Guest (Customer)
[01:04:52] so the next ordered pairs would be (34, 10) and (32, 15)/
[01:04:55] *?

Kathryn C (Tutor)
[01:05:05] yes

Guest (Customer)
[01:06:04] I got 5/-2 again

Kathryn C (Tutor)
[01:06:10] yes

Guest (Customer)
[01:06:52] so the equation is y = 5/-2 + 5? or minus 5?

Kathryn C (Tutor)
[01:10:08] I'm working on this...
[01:11:54] Sorry, I was checking my work and I told you wrong.

Guest (Customer)
[01:12:08] that's okay
[01:12:20] what do we have to fix?

Kathryn C (Tutor)
[01:12:23] Let me correct myself.

Guest (Customer)
[01:12:27] okay

Kathryn C (Tutor)
[01:14:00] I had them backwards. The wind speed is your x values and temp is y values.

Guest (Customer)
[01:14:41] okay

Kathryn C (Tutor)
[01:14:49] So your slope will be -2/5

Guest (Customer)
[01:15:35] and the equation would be y = -2/5 + 5?

Kathryn C (Tutor)
[01:16:08] the 5 is not right. Let me put this on the board, ok?

Guest (Customer)
[01:16:13] okay

Kathryn C (Tutor)
[01:17:19] In this one you will have to find b.
[01:17:25] Can you do that?

Guest (Customer)
[01:18:49] no..

Kathryn C (Tutor)
[01:18:56] ok
[01:19:44] I used your first ordered pair for x and y
[01:20:00] Sorry, wrong again.

Guest (Customer)
[01:20:06] that' alright
[01:21:37] I'm following...

Kathryn C (Tutor)
[01:22:05] 36 = -2/5(5) + b

Guest (Customer)
[01:22:53] -2/5(5) = -2..

Kathryn C (Tutor)
[01:23:06] yes

Guest (Customer)
[01:23:43] 5 = -2 + b

Kathryn C (Tutor)
[01:24:00] 36 = -2 + b

Guest (Customer)
[01:24:20] sorry, I was looking at the board
[01:24:31] how do we find b?

Kathryn C (Tutor)
[01:25:01] add 2 to both sides
[01:25:58] 36 = -2 + b
[01:26:04] add 2 to both sides
[01:27:28] Solve for b

Guest (Customer)
[01:28:00] 38 = 0 + b???

Kathryn C (Tutor)
[01:28:18] yes b = 38
[01:28:41] So your equation would be y = mx + b
[01:29:05] Fill in for m and b

Guest (Customer)
[01:29:05] ohhh. y = -2/5 + 38 ?

Kathryn C (Tutor)
[01:29:21] y = -2/5 x + b
[01:29:37] y = -2/5 x + 38

Guest (Customer)
[01:29:57] okay
[01:30:36] part b: what does the slope mean in this situation?
[01:30:56] it means the temp is going down by -2/5?

Kathryn C (Tutor)
[01:32:38] The temp. is decreasing 2 degrees for every 5mi/hr increase in wind speed.
[01:33:12] Does that make sense?

Guest (Customer)
[01:33:20] I got it! Thanks!

Kathryn C (Tutor)
[01:33:29] ok

Guest (Customer)
[01:34:23] part c: what is the unknown temperature?

Kathryn C (Tutor)
[01:35:06] The temperature is represented by y.

Guest (Customer)
[01:36:41] so the unknown temperature is the actual temp? not just "feels like"?

Kathryn C (Tutor)
[01:37:23] Let me read your problem again

Guest (Customer)
[01:37:57] part c? or the explanation?
[01:38:32] there is a linear relationship between the wind speed at a given temperature and what that temp "feels like". A higher wind speed will make the temperature feel colder. The table below shows what an unknown temp (t) "feels like" at different wind speeds

Kathryn C (Tutor)
[01:39:46] y is what it feels like
[01:40:38] the unknown is the actual

Guest (Customer)
[01:41:19] okay
[01:42:35] and part d says: determine what the unknown temp feels like when the wind speed is 12 mi/hr

Kathryn C (Tutor)
[01:43:15] Put 12 in your equation for x and solve for y

Guest (Customer)
[01:43:22] ok
[01:44:02] in y = mx + b?

Kathryn C (Tutor)
[01:44:28] in y = -2/5x
[01:44:32] sorry
[01:44:46] y= -2/5 x + 38

Guest (Customer)
[01:44:51] oh okay!
[01:46:01] I got y = -4.8x + 38

Kathryn C (Tutor)
[01:46:28] y = -4.8 + 38

Guest (Customer)
[01:46:54] no x?

Kathryn C (Tutor)
[01:47:36] Didn't you put 12 in for x and mult. by -2/5 to -4.8?

Guest (Customer)
[01:48:22] yes. sorry. so is that the final answer? or what would the temp be?

Kathryn C (Tutor)
[01:48:45] -4.8 + 38 = ?

Guest (Customer)
[01:49:55] 33.2

Kathryn C (Tutor)
[01:50:04] yes.

Guest (Customer)
[01:51:19] That is all I have. Thank you so much for the help!

Kathryn C (Tutor)
[01:51:25] Now look at your info. 10 mi/hr ==34 degrees, 15 mi/hr==32 so 12 mi/hr would be between 34 and 32
[01:51:41] and your answer is

Guest (Customer)
[01:51:49] so 33?

Kathryn C (Tutor)
[01:51:57] 33.2

Guest (Customer)
[01:52:10] I got that.

Kathryn C (Tutor)
[01:53:19] The dependent amount is y and the independent is x.
[01:53:48] I'm sorry I got that backwards in the beginning.

Guest (Customer)
[01:54:38] so what does that change?

Kathryn C (Tutor)
[01:55:45] Nothing we did. We caught the mistake with the slope. I just wanted to make sure you knew which one was which.

Guest (Customer)
[01:56:19] Oh, alright. that is all I have.

Kathryn C (Tutor)
[01:57:02] ok. Well I certainly enjoyed working with you. You hung in and did very well. Keep up the good work.

Guest (Customer)
[01:57:27] Thank you for all the help! good night

Kathryn C (Tutor)
[01:57:34] Good night.