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July 09, 2020, 10:53:08 am

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dhpriya

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Help please?
« on: June 05, 2020, 10:34:03 pm »
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Could someone please help me with this question, please?

A car travels half the distance of a journey at an average speed of 80 km/h and a half at an average speed of x km/h. Define a function, S, which gives the average speed for the total journey as a function of x.

keltingmeith

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Re: Help please?
« Reply #1 on: June 05, 2020, 10:54:19 pm »
+1
Sure, what's your problem with it?
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a weaponized ikea chair

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Re: Help please?
« Reply #2 on: June 06, 2020, 12:13:21 pm »
+3
see attachment

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dhpriya

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Re: Help please?
« Reply #3 on: June 06, 2020, 01:01:54 pm »
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Thank you so much I get it now
« Last Edit: June 07, 2020, 11:08:31 am by dhpriya »

dhpriya

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Re: Help please?
« Reply #4 on: June 06, 2020, 01:57:07 pm »
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Could someone also help me with these two questions as well, please?

« Last Edit: June 06, 2020, 02:00:54 pm by dhpriya »

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Re: Help please?
« Reply #5 on: June 06, 2020, 02:18:47 pm »
+2
can't do the first image.

for second one,

f^-1(x)= (b-dx)/(cx-a)

for last part, I think it would be R \ {-d/c , a/c}

Not sure

what is the answer?
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dhpriya

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Re: Help please?
« Reply #6 on: June 06, 2020, 07:11:51 pm »
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The answer to the first question is the first pic and the answer to the second question is the second pic.
I looked at the worked solutions but I didn't quite get it


« Last Edit: June 06, 2020, 07:13:29 pm by dhpriya »

keltingmeith

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Re: Help please?
« Reply #7 on: June 06, 2020, 07:36:43 pm »
+2
The answer to the first question is the first pic and the answer to the second question is the second pic.
I looked at the worked solutions but I didn't quite get it




What don't you get about them? Just showing you the answer wrong fix your understanding, it'll just help you answer this specific question. If you tell me what's bothering you, I can hopefully fill the gaps in your knowledge
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dhpriya

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Re: Help please?
« Reply #8 on: June 06, 2020, 08:18:40 pm »
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What don't you get about them? Just showing you the answer wrong fix your understanding, it'll just help you answer this specific question. If you tell me what's bothering you, I can hopefully fill the gaps in your knowledge


For the first one, I don't understand the graph that they have drawn and for the second question, there is a lot going on so I am confused on how they got the domain of f inverse to be a/c which was before d/c for f and the part after that on the range of f note?

keltingmeith

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Re: Help please?
« Reply #9 on: June 06, 2020, 09:03:49 pm »
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Okay, so the graph they've drawn is for the function - don't even think about t or the shading yet, build things up little by little.

Note that it's a hybrid, or piece-wise, function. So, to draw the graph, you start by drawing the first part. Draw the graph of y=3x. Ignore everything else, just draw that graph. Now, draw the graph of y=3 - that's the second part of our function. Now that you've done this, look at the domain of each part. The first part, y=3x, is only defined for when 0<=x<=1. So, whenever x ISN'T between 0 and 1, erase the line y=3x. That's what gives you the first part of the graph, where is a rising line. For the next part, y=3, is only defined when x>1 - so erase every part of the graph where x ISN'T greater than 1. That's what gives you the flat line in their graph.

Now that you've drawn the piece-wise function, it's time to get to the actual question - what's the area under the curve up to the line x=t? Well, remember those questions in year 10 measurement, where they'd give you those weird shapes you didn't have a formula for? And you had to calculate the area by cutting up the weird shape into parts?

For example, say you have an L-shaped block. To calculate the area, you want to cut the | from the _ (L=|_, can you see it?). Now, it's not the area of a shape you don't have the formula for - it's the area of two rectangles!

Well, we're going to do the same thing here - the total area is going to be the area of the triangle in the first part of the function (y=3x) and the area of the rectangle after that (y=3). So, what's the area of this?

Well, the area of the triangle is (1/2)*b*h. The area of the rectangle is l*w. The question is, what are all these values? You should be able to find them by looking at your graph.

Now, why did they draw two with two different shadings? Well - if the value of t is less than 1, then the line won't include the rectangle, so it'll just be the area of the triangle. Again, you should be able to figure out this area by looking at your graph. Remember, we don't know WHAT value t is - just that at this point, it's less than 1. I encourage you to try playing around with the graph now that you (hopefully) understand it to try and calculate the area for some unknown value of t
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Re: Help please?
« Reply #10 on: June 06, 2020, 09:04:15 pm »
+4

For the first one, I don't understand the graph that they have drawn and for the second question, there is a lot going on so I am confused on how they got the domain of f inverse to be a/c which was before d/c for f and the part after that on the range of f note?

Draw the two graphs first. You should get something that looks like this (https://prnt.sc/suv2y0)

You have no information on what t equals except that it's bigger than 0. It could be between 0 and 1, or bigger than 1. That's why there are two possibilities given in the solutions. If t is between 0 and 1, then you only need to work out the area of a triangle. If bigger than one, then the area would be the triangle underneath f(x)=3x and the rectangle enclosed by x=1, y=3, and x=t.

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dhpriya

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Re: Help please?
« Reply #11 on: June 07, 2020, 11:23:45 am »
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Thank You I totally get it now!

I was just wondering for the second question how they got the domain to be a/c? Could someone please help me with that?

keltingmeith

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Re: Help please?
« Reply #12 on: June 07, 2020, 08:17:38 pm »
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Thank You I totally get it now!

I was just wondering for the second question how they got the domain to be a/c? Could someone please help me with that?

Whoops, I knew I forgot something.

Remember - for an inverse function, you swap all of x and all of y. It's like taking the graph the paper is on and turning it over diagonally so you're now looking at it from the back. OF COURSE the domain might change of you do something so drastic, don't assume that it won't. However, if all of x swaps with all of y, and the domain represents all the x values, what would you expect the new domain to be?
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dhpriya

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Re: Help please?
« Reply #13 on: June 07, 2020, 08:38:52 pm »
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Whoops, I knew I forgot something.

Remember - for an inverse function, you swap all of x and all of y. It's like taking the graph the paper is on and turning it over diagonally so you're now looking at it from the back. OF COURSE the domain might change of you do something so drastic, don't assume that it won't. However, if all of x swaps with all of y, and the domain represents all the x values, what would you expect the new domain to be?

OHH OK I get it.
Thank you