query 5

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course Mth 173

8:13 p.m. 2/4

005. `query 5

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Question: `q Growth rate and growth factor: Describe the difference between growth rate and growth factor and give a short example of how each might be used

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Your solution:

growth rate is the percentage of change the 10% is a growth rate of .1, growth factor is the total amount, with 10% it would be 1 (the base, original value) plus the percent of change .1, to make 1.1.

if you start with $100 dollars, and the next month you received 110, you know the growth rate was .1, if you wanted to project that farther, you would use the growth factor.

confidence rating #$&*:2

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Given Solution:

** Specific statements:

When multiplied by a quantity the growth rate tells us how much the quantity will change over a single period.

When multiplied by the quantity the growth factor gives us the new quantity at the end of the next period. **

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Self-critique (if necessary):OK

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Self-critique Rating:OK

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Question: `q Class notes #05 trapezoidal representation.

Explain why the slope of a depth vs. time trapezoid represents the average rate of change of the depth with respect to the time during the time interval represented

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Your solution:

On a trapezoid the line will be straight for the slope, and not all data is a perfect straight line, so that is an average, not an actual amount, however the smaller, and because of that more, that you have the closer it will be to actual.

confidence rating #$&*:3

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Given Solution:

** GOOD ANSWER BY STUDENT WITH INSTRUCTOR COMMENTS:

The slope of the trapezoids will indicate rise over run

or the slope will represent a change in depth / time interval

thus an average rate of change of depth with respect to time

INSTRUCTOR COMMENTS:

More detail follows:

** To explain the meaning of the slope you have to reason the question out in terms of rise and run and slope.

For this example rise represents change in depth and run represent change in clock time; rise / run therefore represents change in depth divided by change in clock time, which is the average rate of change. **

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Self-critique (if necessary):

my answer seems to vary slightly, but I do understand this

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Self-critique Rating:3

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Question: `q Explain why the area of a rate vs. time trapezoid for a given time interval represents the change in the quantity corresponding to that time interval.

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Your solution:

Im not sure I understand the question, it sounds to me like it wants to know how one point of the slope on the trapezoid and the distance to the other one represent the amount of change between the two, I do not know how to express this in terms of the

trapezoid. but the distance from point 1 to 2 is the change, and if you draw a straight line down from those points, the area between them is the change.

confidence rating #$&*:1

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Given Solution:

**STUDENT RESPONSE WITH INSTRUCTOR COMMENTS:

The area of a rate vs. time graph rep. the change in quantity.

Calculating the area under the graph is basically integration

The accumulated area of all the trapezoids for a range will give us thetotal change in quantity.

The more trapezoids used the more accurate the approx.

INSTRUCTOR COMMENTS: All very good but the other key point is that the average altitude represents the average rate, which when multiplied by the width which represents time interval gives the change in quantity

You have to reason this out in terms of altitudes, widths and areas.

For the rate of depth change example altitude represents rate of depth change so average altitude represents average rate of depth change, and width represents change in clock time.

average altitude * width therefore represents ave rate of depth change * duration of time interval = change in depth.

For the rate of change of a quantity other than depth, the reasoning is identical except you'll be talking about something besides depth. **

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Self-critique (if necessary):

I understand what is happening, I just don't get it in the terms you are saying, altitudes, widths and areas. Let me try to word what I understand in by looking at it as a trapezoid (it kind of confuses me to look at it in this way, but I will try).

the top of the trapezoid is a slop (average).

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Most specifically the top is a sloping line segment. The slope is the slope of that segment.

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the the y-axis therefore is representing an altitude, at an average, since any point we pick will be on this slope, which is an average.

so picking two points on this, and drawing a line straight down, the area inside, will be the area.

if you only measure the x-axis of these two points, you have the width.

since the slope is an average, each tick you go across on the x-axis is representing an AVERAGE change, since the slope is a change.

and we basically have described a rise over run.

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On a graph of rate vs. time, the vertical quantity, the quantity measured relative to what we often call the y axis, is a rate of change (presumably a rate of change of some quantity with respect to time).

The sloping segment at the top of the trapezoid is part of the graph of the rate vs. clock time. The 'graph altitudes' are the distances from the x axis to the points of the graph, i.e., from the x axis to the sloping top of the trapezoid.

These distances are measured relative to the y axis, so they are rates. Thus the 'graph altitudes' of the trapezoid represent rates.

The altitude in the middle of the trapezoid is the average altitude, and therefore represents the average rate.

The width of the trapezoid is measured along the horizontal axis, what we often think of as the x axis. In this case we would probably call the horizontal axis the t axis, since clock time is measured along this axis.

The width therefore represents a change in clock time.

When we multiply an average rate of change with respect to time by the corresonding change in clock time, we get the change in the quantity, since

average rate of change of quantity with respect to clock time = (change in quantity) / (change in clock time).

Rise and run have nothing to do with the calculation of the area of the trapezoid. Only its average 'graph altitude' and width affect its area.

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Self-critique Rating:1

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Question: `q ÿÿÿ #17. At 10:00 a.m. a certain individual has 550 mg of penicillin in her bloodstream. Every hour, 11% of the penicillin present at the beginning of the hour is removed by the end of the hour. What is the function Q(t)?

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Your solution:

well the growth rate is decreasing, so instead of .11 we should make it -.11

550(1-.11)^t

where t is the time in hours.

confidence rating #$&*:2

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Given Solution:

** Every hour 11% or .11 of the total is lost so the growth rate is -.11 and the growth factor is 1 + r = 1 + (-.11) = .89 and we have

Q(t) = Q0 * (1 + r)^t = 550 mg (.89)^t or

Q(t)=550(.89)^tÿ **

How much antibiotic is present at 3:00 p.m.?

** 3:00 p.m. is 5 hours after the initial time so at that time there will be

Q(5) = 550 mg * .89^5 = 307.123mg

in the blood **

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Self-critique (if necessary):

I did not see a part asking me to evaluate it for a certain time.

???is it necessarily wrong for me to leave it as 550(1-.11)^t. I think this is a more pleasurable response than simplifying.

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Before you do any repeated calculations with this function you'll probably want to simplify it.

However leaving it in the form you prefer, which shows more clearly where the expression comes from and how it is related to the conditions of the problem, definitely has its advantages and if you prefer that form, by all means use it.

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Self-critique Rating:OK

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Question: `q Describe your graph and explain how it was used to estimate half-life.

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Your solution:

what graph? I don't recall having a graph, but I can imagine one and attempt to descripe it.

half-life is the amount of time it takes for the radioactive to be halved. so if you graph out the curve that represents half-life, you would pick one point, and travel down to the point where y is half its orignal value, and find the change in x, and that is its half-life. It

doesnt matter what point you pick, it will always be the same amount of time change

confidence rating #$&*:2

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Given Solution:

** Starting from any point on the graph we first project horizontally and vertically to the coordinate axes to obtain the coordinates of that point.

The vertical coordinate represents the quantity Q(t), so we find the point on the vertical axis which is half as high as the vertical coordinate of our starting point. We then draw a horizontal line directly over to the graph, and project this point down.

The horizontal distance from the first point to the second will be the distance on the t axis between the two vertical projection lines. **

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Self-critique (if necessary):OK

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Self-critique Rating:OK

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Question: `q What is the equation to find the half-life?ÿ What is its most simplified form?

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Your solution:

would this be the same as doubling time.?

2P=P(1+r)^(t+'dt)-P(1+r)^2

I still have a little difficulty with this equation

I know the simplist would be P^doublingTime=2p, which could be used for half time too I am assuming.

confidence rating #$&*:1

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Given Solution:

** Q(doublingTime) = 1/2 Q(0)or

550 mg * .89^doublingTIme = .5 * 550 mg. Dividing thru by the 550 mg we have

.89^doublingTime = .5.

We can use trial and error to find an approximate value for doublingTIme (later we use logarithms to get the accurate solution). **

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Self-critique (if necessary):OK

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Self-critique Rating:OK

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Question: `q #19. For the function Q(t) = Q0 (1.1^ t), a value of t such that Q(t) lies between .05 Q0 and .1 Q0.

!!!i believe the template is a little messed up, and there is a self critque separating this and the problem, and confused me for a second.

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Self-critique (if necessary):

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Question: `q For what values of t did Q(t) lie between .005 Q0 and .01 Q0?

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Your solution:

For the function Q(t) = Q0 (1.1^ t), a value of t such that Q(t) lies between .05 Q0 and .1 Q0.

if the rate is increasing then t would have to be negative right?

i used a principle of 200, so using tril and error i am finding ranges from 10-20 by plugging in for t.

-20 gave almost 30

-25 gives 18, so that number is to high

-23 gives a little over 22

-24 gives a little over 20

-24.158 gives 20.00163545

-24.159 gives 19.99978231 I will choose to say -24.159 for .1 of Q0

-30 gives a little over 11

-31 gives a little over 10

-31.431 gives a little over 10... 10.00038013 so i will use this for .05

so for the range it is about -31.431 to -24.159, of course I didnt get to close so the numbers will be slightly off.

confidence rating #$&*: 2

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Given Solution:

** Any value between about t = -24.2 and t = -31.4 will result in Q(t) between .05 Q0 and .1 Q0.

Note that these values must be negative, since positive powers of 1.1 are all greater than 1, resulting in values of Q which are greater than Q0.

Solving Q(t) = .05 Q0 we rewrite this as

Q0 * 1.1^t = .05 Q0. Dividing both sides by Q0 we get

1.1^t = .05. We can use trial and error (or if you know how to use them logarithms) to approximate the solution. We get

t = -31.4 approx.

Solving Q(t) = .1 Q0 we rewrite this as

Q0 * 1.1^t = .1 Q0. Dividing both sides by Q0 we get

1.1^t = .1. We can use trial and error (or if you know how to use them logarithms) to approximate the solution. We get

t = -24.2 approx.

(The solution for .005 Q0 is about -55.6, for .01 is about -48.3

For this solution any value between about t = -48.3 and t = -55.6 will work). **

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Self-critique (if necessary): OK

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Self-critique Rating:OK

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Question: `q explain why the negative t axis is a horizontal asymptote for this function.

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Your solution:

whatever number you put in, you can get really close to zero, but never reach it (on the y-axis) and looking in real life situations this makes more sense, especially when halving things.

confidence rating #$&*:2

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Given Solution:

** The value of 1.1^t increases for increasing t; as t approaches infinity 1.1^t also approaches infinity. Since 1.1^-t = 1 / 1.1^t, we see that for increasingly large negative values of t the value of 1.1^t will be smaller and smaller, and would in fact approach zero.

**

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Self-critique (if necessary):

I didn't think about the infinity side of it, I dont see its purpose in explaining the horizontal asymptote, except maybe that we are using a negative exponent which is the same as putting it into the denominator.

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Self-critique Rating:2

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To see that 1 / 1.1^t approaches zero, it's very useful to understand that 1.1^t approaches infinity.

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Question: `q #22. What value of b would we use to express various functions in the form y = A b^x? What is b for the function y = 12 ( e^(-.5x) )?

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Your solution:

The principle?

I don't remember the exact value for e, but I have a calculator that has e, and when i plug it in e^-.5 = .60653 rounded.

that would turn the equation to 12* .60653^x (im not sure how we seperate the x, I just remember we can)

so b = .60653

confidence rating #$&*:

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Given Solution:

** 12 e^(-.5 x) = 12 (e^-.5)^x = 12 * .61^x, approx.

So this function is of the form y = A b^x for b = .61 approx.. **

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Self-critique (if necessary):OK

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Self-critique Rating:OK

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Question: `q what is b for the function y = .007 ( e^(.71 x) )?

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Your solution:

would this not be the same as the last one in the way we would do it

.007 * 2.03399^x

so b= 2.03399 rounded

confidence rating #$&*: 2

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Given Solution:

** .007 e^(.71 x) = .007 (e^.71)^x = .007 * 2.04^x, approx.

So this function is of the form y = A b^x for b = 2.041 approx.. **

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Self-critique (if necessary):OK

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Self-critique Rating: OK

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Question: `q what is b for the function y = -13 ( e^(3.9 x) )?

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Your solution:

-13 * 49.402449^x

b= 49.402449 rounded

confidence rating #$&*:3

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Given Solution:

** -13 e^(3.9 x) = -13 (e^3.9)^x = -13 * 49.4^x, approx.

So this function is of the form y = A b^x for b = 49.4 approx.. **

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Self-critique (if necessary): OK

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Self-critique Rating:OK

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Question: `q List these functions, each in the form y = A b^x.

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Your solution:

y=2* .60653^x

y=.007 * 2.03399^x

y=-13 * 49.402449^x

confidence rating #$&*:3

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Given Solution:

** The functions are

y=12(.6065^x)

y=.007(2.03399^x) and

y=-13(49.40244^x) **

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Self-critique (if necessary):OK

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Self-critique Rating:OK

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Question: `q query text problem 1.1.31 5th; 1.1.23 4th dolphin energy prop cube of vel

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Your solution:

what is this? I have no idea what this is talking about, maybe i skipped over something?

confidence rating #$&*: 0

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Given Solution:

** A proportionality to the cube would be E = k v^3. **

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Self-critique (if necessary):

I am completely lost. if this is from the text_05 link in the brief assignments page, this link is broken for me, and does not redirect me.

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Self-critique Rating:

@&

The links work on the full Assignment Page (at least the link to Text_05 works).

I'll check the rest and correct them as needed, so by the time you read this they should all be working.

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Question: `q query text problem 1.1.37 5th; 1.1.32 4th temperature function H = f(t), meaning of H(30)=10, interpret vertical and horizontal intercepts

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Your solution:

again, I am completely lost.

confidence rating #$&*:0

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Given Solution:

** The interpretation would be that the vertical intercept represents the temperature at clock time t = 0, while the horizontal intercept represents the clock time at which the temperature reaches zero. **

what is the meaning of the equation H(30) = 10?

** This means that when clock time t is 30, the temperature H is 10. **

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Self-critique (if necessary):

If I need to do these, could you copy the questions down so I can try to solve them, I do not know where these are pulled from

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Self-critique Rating:

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Question: `q What is the meaning of the vertical intercept?

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Your solution:

the vertical intercept is the y-intercept, it means where is the graph when x=0, which should be touching the ""bar"" of the y-axis. (if the graph exists at this point)

confidence rating #$&*:

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Given Solution:

** This is the value of H when t = 0--i.e., the temperature at clock time 0. **

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Self-critique (if necessary):

Even though I don't know what the problem is, I still think I understood what the intercepts are

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Self-critique Rating:

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Question: `q What is the meaning of the horizontal intercept?

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Your solution:

this is the x-intercept, the point of the graph when y= 0, and normally touches the ""bar"" of the x-axis, if it exists at this point.

confidence rating #$&*:

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Given Solution:

** This is the t value when H = 0--the clock time when temperature reaches 0 **

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Self-critique (if necessary):OK

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Self-critique Rating:OK

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Question: `q query text problem 1.1.40 5th; 1.1.31 4th. Water freezes 0 C, 32 F; boils 100 C, 212 F. Give your solution.

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Your solution:

is this asking for the conversion of celsius to fahrenheit (i don't remember how to spell it). If so i cant remember the formula, i believe it was something/(C+32) i think. but if this is what it is looking for I may be able to figure this one out.

looking at what I remeber it means it is setup as (212-32)/(100-0) I just noticed, the equation i remember is + in the denominator, slope is a negative...so something is different. 180/100 9/5... trying to figure out something. Oh right I just found the slope the line

can be shifted so I need an equation

F=9/5x+b

in our points x were our celsius points....so at 100C. 212=(9/5)100+b

212=180+b I noticed this is what I had in my numerator of slope, is this coincidence, or something to it?

212-180=b

32=b AHA thats there that 32 I remember came from.

so F=(9/5)C+32

and to solve for C i remember it was just like, a recipricole (or something like that?)

so C= (5/9)F-32 and I can't remember why the sign changes, just my logic is if F is C+32 then thats going up, so if you are at c going to f, you will be going down like if x to y is climbing, then y to x will be falling.

confidence rating #$&*:1

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Given Solution:

** The graph contains points (0, 32) and (100, 212). The slope is therefore (212-32) / (100-0) = 1.8.

The y-intercept is 32 so the equation of the line is

y = 1.8 x + 32, or using F and C

F = 1.8 C + 32.

To find the Fahrenheit temp corresponding to 20 C we substitute C = 20 into F = 1.8 C + 32 to get

F = 1.8 * 20 + 32 = 36 + 32 = 68

The two temperatures will be equal when F = C. Substituting C for F in F = 1.8 C + 32 we get

C = 1.8 C + 32. Subtracting 1.8 C from both sides we have

-.8 C = 32 or

C = 32 / (-.8) = -40.

The scales read the same at -40 degrees. **

STUDENT QUESTION

I understand all work shown except the answer for d? I understand how to substaute and see how you worked it, but don’t understand the logic for why this works ?

INSTRUCTOR RESPONSE

You have a linear function which gives you F when you substitute C.

That means that the graph of F vs. C is a straight line.

According to the given information the straight line passes through the points (0, 32) and (100, 212).

Given two points, there are a variety of ways to get the equation of the corresponding straight line. The given solution finds the slope, then uses slope-intercept form of the equation of a straight line.

Alternatively you could use the point-point form of the equation, which would lead to the same result.

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Self-critique (if necessary):

I don't understand the part of scales reading the same at -40 degrees.

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Self-critique Rating:1

@&

The Fahrenheit scale is higher than the Celsius scale for temperatures greater than 0, and for many temperatures lower than zero. However since Celsius degrees are bigger than Fahrenheit degrees, as temperatures lower the Fahrenheit scale lowers at a greater rate than the Celsius scale. So the two must intersect at some point.

If you graph F = 9/5 C + 32 vs. C you will get a graph with vertical intercept 32 and slope 9/5.

If you graph C vs. C you will get a line through the origin with slope 1.

These lines intersect at C = -40.

At this point then 9/5 C + 32 is equal to C, both being equal to -40.

You get the same result if you set the y coordinate of the graph of F = 9/5 C + 32 equal to C. The equation is 9/5 C + 32 = C, which says that the Fahrenheit temperature 95 C + 32 is equal to the Celsius temperature C.

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Self-critique (if necessary):

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Self-critique rating:

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Self-critique (if necessary):

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Self-critique rating:

#*&!

@&

The Text_ links on the full Assignments Page worked, but those on the Brief Assignments Page were accidentally linked to a folder on the host computer. I've corrected that, I think. Unfortunately my browser (Chrome) doesn't allow me to see recent changes so I can't double-check that tonight, so let me know if it doesn't work.

*@

&#This looks good. See my notes. Let me know if you have any questions. &#