You are not logged in.
hi tammybrown
Welcome to the forum.
Oh bliss! You have asked about one of my favourite topics and I've got more than one post's worth to tell.
Let's start with snowdrops. They grow in woods; coming up usually early in the year; flower; then die back and disappear from the woodland floor. But, next year, up they come again; mostly growing from the bulbs they made the previous year. They can grow from seed; that's how they spread around the wood; but they also spread by budding off a second bulb from the first.
There are biological rules for this. A seed may grow that year then all the nutrients in the leaves are sucked back into a bulb which acts as a store for fresh growth next year. Each bulb produces a single plant stem and flower so it's easy to count the bulbs without having to dig them up. A first year bulb has only enough nutrients to produce a plant stem and flower but a mature bulb (ie. more than one year old) can also bud off a new first year bulb.
So at the end of year 1 the plant makes a first year bulb for next year. At the end of the second year that bulb gets 'promoted' to a mature bulb and so produces a new first year bulb. That's two bulbs for the third year.
At the end of the third year the mature one produces another first year bulb and the first year bulb gets promoted. So we have two matures and one first year.
At the end of the fourth year the two matures produce two first years, and the first year gets promoted. That's three matures and two first years.
I'll put all of this into a table:
end of year firsts matures
1 1 0
2 1 1
3 1 2
4 2 3
5 3 5
6 5 8
7 8 13
..................................................................
It doesn't always work because an animal may eat a bulb or a human may step on a plant and kill it but the majority of clumps have a number of flowers(=bulbs) in the Fibonacci sequence. When a disaster occurs and the clump loses a bulb the rule still applies but maybe not with the familiar numbers eg 1 1 2 3 disaster one lost 2 remain, 4,6,10,16 etc.
Here's some homework for you to try. If you can do it on a spreadsheet that will save you a lot of arithmetic but you can do it on paper with a calculator.
Write out the Fibonacci sequence. Create a new sequence by calculating each number divided by the one before:
eg 1/1 = 1 2/1 = 2 3/2 = 1.5 ..............and so on.
What do you notice about your new sequence? What else about it?
Bob
Bob
hi MathAppreciator
Welcome to the forum.
That post was in 2021 and we've had nothing else from Lorbeer54.
Sometimes a link is not what it appears to be and may have been deleted (not by me) for good reasons.
I fear we'll never know
Later edit. On the other hand I enjoy a bit of detective work and with the help of my friend Google I came up with this:
https://maths-in-story.weebly.com/
It looks harmless to me and I tried a few pages. Let me know if I've made a mistake.
Bob
r(x) = R(x-5) = 170.7(x-5)^2 + 1373(x-5) + 1080
If you put x = 0 that computes R(0-5) = R(-5)
However when you put x = 5 that computes R(5-5) = R(0) which is 1080 the amount for 2005.
If the formula works for years before 2005 then x=0 will compute the revenue for a year 5 years before2005, ie 2000.
Bob
I looked up Shalom and it's a greeting meaning peace. Lovely name!
Bob
0 < x ≤ 100 cost per mile 0.5 max cost = 100 x 0.5 = 50
100 < x ≤ 300 cpm 0.4 max cost = 200 x 0.4 = 80
300 < x ≤ 400 cpm 0.25 max cost = 100 x 0.25 = 25
400 < ≤ 960 cpm no charge
So:
0<x≤100 C = 0.5x
100<x≤300 C = 50 + 0.4(x-100)
300<x≤400 C = (50+80) + 0.25(x-300)
400x≤960 C = (50 + 80 + 25)
I've put the 'already paid' amounts in brackets so you can see where the numbers comes from. You can complete the additions in your answer.
Plot these points: (0,0) (100,50) (300, 130) (400, 155) (960, 155) and join them with straight lines.
Your answers in part (a) tell us the digital revenue (how much money was made) in the years x = 0, 3 and 5 which means the years 2005, 2007, 2009.
r(x) = R(x-5) = 170.7(x-5)^2 + 1373(x-5) + 1080
The (c) answers should be the same as the (a) answers.
In this new model when x = 5 the year is 2005, so the new model is the same as the old except the start year is now 2000.
Bob
Both exactly correct!
Bob
Both answers are ok but there was no need to include "on the line ....." Saying down or right is enough.
Bob
Shalom Shalom,
Welcome to the forum.
Bob
I'm like the Canadian Mounties; I never give up. I'll post more help if you want. Just say.
Bob
Both new functions reflect the original in an axis. So it's the same for both: increasing from -2 to + 7.
Bob
I think so.
It would have been better if the questioner had said "For each function identify if it is decreasing, increasing or neither including where in the domain this occurs." But they possibly thought it was obvious.
Bob
When the red graph is decreasing, the blue graph is increasing.
When the blue graph is increasing, the red graph is decreasing.
Correct.
The graph of y = -f(x) is decreasing at the point (7, 0).
The graph of y = f(x) is increasing at the point (7, 0).
The graph I made up had zero gradient at the end points of the interval.
The original question describes the interval as (-2, 7) . Don't get confused here with coordinates. This doesn't mean x=-2, y = 7; but rather all x values from -2 to +7. As the brackets are round not [ ] the end points are not included so it's ok for me to make up a graph that stops decreasing at the endpoints. If it's not decreasing it must either have an instantaneous change to positive or go through zero gradient at a local maximum or minimum. I chose the latter as it's easier to make up an equation for this.
Bob
Is this right? No. You're charging too much.
For part 2, 100 miles has already been charged so 0.4 only applies for 200 miles (300 minus 100)
Same for the remaining parts. Subtract the miles that have already been charged to determine how many more miles to charge at the next rate.
A step function graph looks like a staircase, flat parts getting higher.
This graph has sloping lines with gradients 0.5, 0.4, 0.25, 0 so the sections go up in sloping lines except for the last which is flat.
Bob
I have made up a function, shown in red, that is decreasing from x = -2 to x = +7
-f(x) is a reflection in the x axis. I have shown that graph in blue.
From x = -2 to x = + 7 the blue graph is increasing.
Bob
A = (-2, -1)
B = (-1, -1)
C = (1, 1)
D = (2, 0).
A. Draw the graph of y = | f(x) |. You have to work out f(x) and then make any negative values into positives. Some are already positive and so those are unchanged.
B. Draw the graph of y = f(| x |) You have to change any negative x values into positives; then use the function to work out what y values you get. I found that the original 4 points become just 2 points repeated.
What is the basic difference? Neither new graph looks anything like the original; nor do they share any similarities. Hard to see what the questioner is searching for here.
What you can say is that y = | f(x) | has no negative y values and y = f(| x |) has no negative x values. Is that what is wanted?
Bob
Correct.
Bob
Part A fully correct.
Looks like a typo crept in for B . (3,12) was correct but then you put (3,10)
Bob
You did the hard bit in post 2.
A graph that is decreasing means it slopes downwards as you go from left to right x=2 to x=7.
If you reflect it what does that do to the slope?
Bob
I've added an extra column to my table:
0 < x ≤ 100 cost per mile 0.5 max cost = 100 x 0.5 = 50
100 < x ≤ 300 cpm 0.4 max cost = 200 x 0.4 = 80
300 < x ≤ 400 cpm 0.25 max cost = 100 x 0.25 = 25
400 < ≤ 960 cpm no charge
C(x) = {0.50x, if 0 ≤ x ≤ 100....Part 1
This looks ok.
C(x) = {50 + 0.40(x - 100), if 100 < x ≤ 400...Part 2
Not x ≤ 400. The upper limit for part 2 is 300.
C(x) = {170 + 0.25(x - 400), if 400 < x ≤ 800...Part 3
Where did 170 come from?
Bob
Yes, that's right.
Bob
Mathematically, this is similar to the truck charge question. First summarise the info.
0 < weight ≤ 1 fixed charge of 1.17
1 < x ≤ 13 0.17 per ounce
So between x=0 and 1 the graph will be a horizontal line as the charge is fixed and doesn't vary with x.
After that the additional cost starts to go up in a straight line with gradient 0.17 .
The graph stops when x=13
Bob
I like to try and summarise the written information using (in this case) algebra.
0 < x ≤ 100 cost per mile 0.5
100 < x ≤ 300 cpm 0.4
300 < x ≤ 400 cpm 0.25
400 < ≤ 960 cpm no charge
But, beware. This doesn't show fully what a charge will be, nor does it give you the points for a graph. Foe example, if the distance for a package is, say, 150 miles then the charge would be 100 x 0.5 + 50 x 0.4
So, to get the function you need to include charges for a previous stage and account for how many more miles have been travelled. I'll show what I mean for distances in the 100 - 300 group.
100 < x ≤ 300 total charge = 100 x 0.5 + (x-100) x 0.4
I'll leave you to deal with the other groups similarly. That should enable you to make the graph.
Bob
That's it. Well done.
Bob