Last Assessment Please Help Me!!

mantastrike · 2011-05-18 14:10:28

Thanks For All!!

Last edited by mantastrike (2011-05-30 19:44:52)

bobbym · 2011-05-18 14:18:46

Hi mantastrike;

Please check problem #2. As it is written it does not make much sense.

mantastrike · 2011-05-18 14:24:06

bobbym wrote:

Hi mantastrike;
Please check problem #2. As it is written it does not make much sense.

Thanks! I'll show you my progress tonight or tomorrow.

bobbym · 2011-05-18 14:29:09

Okay, thanks for rewriting it.

2a)

2b)

5. An athlete aims to improve her time on a given course by 1.5 seconds a day. If on the first day she runs the course in 4.25 minutes, how fast must she run it on the 21st day to stay on target?

Form the recurrence:

Run the recurrence or solve for it. a(21) = 3 minutes and 45 seconds. ( thanks again John E. Franklin )

i. Write out the first four terms of the sequence.
ii. Find an explicit formula.
iii. Use this formula to find the 5th term.
iv. Check that the formula works correctly on the first 5th terms.

Solve using the characteristic equation or generating functions or by unrolling the recurrence out.

i)

3,11,35,107,...

ii)

iii)

iv) This one you check by plugging in n = 0,1,2,3,4 into ii)

8a) No, for a connected graph to have an Euler path it must have 0 or 2 odd degree vertices. Count up the number and that will show more than 2

8b) Yes, it is possible to visit each vertex once and only once. You can prove that easily.

John E. Franklin · 2011-05-18 14:47:43

8. a.) no Euler paths exist because there are too many odd number of edges at many vertices, so when you go in and out, then you are left with in, but then you get stuck there before you finish the other ones.

8. b.) yes, a Hamilton path does exist because it does in a cube, so it does even without the extra curved line. Simply go clockwise around half of the cube and then the other way as viewed through the cube which is the same clockwise if you go to the other side, but it is against the tide of the other one, and then jump back across to the start making 4 + 4 dots.

mantastrike · 2011-05-18 15:16:12

John E. Franklin wrote:

8. a.) no Euler paths exist because there are too many odd number of edges at many vertices, so when you go in and out, then you are left with in, but then you get stuck there before you finish the other ones.
8. b.) yes, a Hamilton path does exist because it does in a cube, so it does even without the extra curved line. Simply go clockwise around half of the cube and then the other way as viewed through the cube which is the same clockwise if you go to the other side, but it is against the tide of the other one, and then jump back across to the start making 4 + 4 dots.

can you please show workings?

John E. Franklin · 2011-05-18 15:16:40

I'm guessing here, but maybe I'm right?!
3. a.)
threeplace Boolean Function F(x,y,z) = (3x + 2y + z) mod 2.
xyz=F
000=0
001=1
010=0
011=1
100=1
101=0
110=1
111=0
This is just guesswork by plugging in zeros and ones.
As for part b.), I don't know what the arrows are for yet.

John E. Franklin · 2011-05-18 15:27:38

Here is
a picture
of 8. b. )

Click to enlarge.

Notice I redrew the graph into 3-D, but
it is the same graph.

John E. Franklin · 2011-05-18 15:48:08

On #5, I get a slightly different answer because I am using elementary school techniques.
Day 21 is 20 days later than day #1, so 20 times 1.5 is 30seconds less.
So 3.75 minutes is 3:45 minutes versus start of 4:15 minutes.
The velocity or speed in a straight line run without weaving would have to
increase from the original speed, called "initial speed" or IntlSpd, to a
greater speed with the simple ratio or factor multiplier of 4.25 over 3.75 times
the IntlSpd. So 21st day speed is (425/375)IntlSpd, or (85/75)IntlSpd, or
(17/15)IntlSpd. [I'm not sure why bobby has a more complicated reasoning?!,
but I'm sure there is a reason I just don't know.]

bobbym · 2011-05-18 16:03:49

Hi John;

There is a good reason for the discrepancy in answers. Mine is a typo. First I typed 1.5 and then for some reason I typed 1.25. Your answer of 225 is correct.

The more complicated reasoning, that I have an answer for. Question 6 is all about recurrences so I wanted to start him thinking that way. I thought if he saw one used in #5 he would be better able to answer his own questions in #6.

Anyway, thanks for spotting that, I will make the necessary changes to the above post so as to avoid confusion. Sorry for the confusion.

mantastrike · 2011-05-23 21:17:07

Question 4. I used formula sum of vertex degrees = 2x no. of edges which means 11x2 = 22

then the sum of vertex degrees is = 1+2+3+4+3+5 = 18

therefore 18+x = 2x11

= 18 - 18 + x = 22 - 18

x = 3

Did I got it right??

mantastrike · 2011-05-23 21:21:36

bobbym wrote:

Okay, thanks for rewriting it.
2a)
2b)

5. An athlete aims to improve her time on a given course by 1.5 seconds a day. If on the first day she runs the course in 4.25 minutes, how fast must she run it on the 21st day to stay on target?
Form the recurrence:
Run the recurrence or solve for it. a(21) = 3 minutes and 45 seconds. ( thanks again John E. Franklin )
i. Write out the first four terms of the sequence.
ii. Find an explicit formula.
iii. Use this formula to find the 5th term.
iv. Check that the formula works correctly on the first 5th terms.
Solve using the characteristic equation or generating functions or by unrolling the recurrence out.
i)
3,11,35,107,...
ii)
iii)
iv) This one you check by plugging in n = 0,1,2,3,4 into ii)
8a) No, for a connected graph to have an Euler path it must have 0 or 2 odd degree vertices. Count up the number and that will show more than 2
8b) Yes, it is possible to visit each vertex once and only once. You can prove that easily.

Please tell me, are these exact answers and can just copy it? or Do I still have to figure it out.

bobbym · 2011-05-23 21:27:19

Hi;

I never recommend just copying answers. In math we can check things for ourselves. Yes, it is what I would put down.

iv) You have to plug in.

mantastrike · 2011-05-23 22:32:11

bobbym wrote:

Hi;
I never recommend just copying answers. In math we can check things for ourselves. Yes, it is what I would put down.
iv) You have to plug in.

Plug-in? how do I plug-in? sorry english is not my first language. Can you please tell me exactly what should I do? and also can you please check if I got right answer for question 4.

bobbym · 2011-05-23 22:40:53

Hi;

4)

You are using the right formula but your algebra is incorrect.

= 18 - 18 + x = 22 - 18

x = 3

It is supposed to be x = 4

6)

From the recurrence you get the first 6 terms.

3, 11, 35, 107, 323, 971...

iv) To plug in the formula you replace n with the numbers 0,1,2,3,4. You should get the above numbers.

First one checks.

Second one checks.

Third one checks.

Can you do the remaining two?

mantastrike · 2011-05-23 23:49:49

John E. Franklin wrote:

I'm guessing here, but maybe I'm right?!
3. a.)
threeplace Boolean Function F(x,y,z) = (3x + 2y + z) mod 2.
xyz=F
000=0
001=1
010=0
011=1
100=1
101=0
110=1
111=0
This is just guesswork by plugging in zeros and ones.
As for part b.), I don't know what the arrows are for yet.

Please help me with question 3.

mantastrike · 2011-05-24 00:00:34

bobbym wrote:

Hi;
4)
You are using the right formula but your algebra is incorrect.

= 18 - 18 + x = 22 - 18
x = 3
It is supposed to be x = 4

6)
From the recurrence you get the first 6 terms.
3, 11, 35, 107, 323, 971...
iv) To plug in the formula you replace n with the numbers 0,1,2,3,4. You should get the above numbers.
First one checks.
Second one checks.
Third one checks.
Can you do the remaining two?

Thanks boby, can you please help me with the remaining. You don't have to worry about no. 7 I would still show you my answer for you to check. This is kinda frustrating because most of the questions given are not discussed in the class. I was only able to answer questions that are.
Please help me I just want to finish this course, aside from this I have other assessments that I am finishing that is why I need assistance from online resources such as this message board in regards to mathematics.

Thank you.

bobbym · 2011-05-24 03:25:47

Hi mantastrike;

aside from this I have other assessments

I do not agree with these type statements. Math is the most important course you are taking. You have to learn it.

Did you try the last two as I suggested?

anonimnystefy · 2011-05-24 05:03:14

i wanted to ask something and found this topic.
i was interested in logarithmic function and got most of it but i don't know how to do problems like this:
logx+2 sqrt(5*x/2+4)>1 where x+2 is the base.
can anyone help me?

bobbym · 2011-05-24 05:11:49

Hi anonimnystefy;

Is this what you mean?

anonimnystefy · 2011-05-24 05:15:33

hi bobbym,

yeah that's it.
i tried something and got that -3/2<x<0, but i'm not sure.

bobbym · 2011-05-24 05:28:21

Hi;

That interval is not correct.

- 5 / 4 is not a solution to that inequality.

Can I see what you did?

anonimnystefy · 2011-05-24 05:47:50

hi bobbym;

x+2 will always be the base just so you don't get confuse;
log(x+2) sqrt(5x/2+4)>1
1/2 log(x+2) (5x+8)/2 >1
log(x+2) (5x+8) - log(x+2) 2 >2
log(x+2) (5x+8) - log(x+2) 2 > log (x+2) sqr(x+2) sqr is 'square of' and the last 'x+2' is not in the base
log(x+2) (5x+8) > log(x+2) 2(sqr(x)+4x+4)
5x+8 > 2sqr(x) +8x+8
2sqr(x)<-3x
now the first case is that x is mre than 0
we divide by x
2x<-3
x<-3/2 but than x is not > 0

the second case is x<0
we divide by x s the sign changes
2x>-3
x>-3/2
x<0
so -3/2<x<0;

the third case is x=0 but then we get 0<0 which is always false

Last edited by anonimnystefy (2011-05-24 05:51:25)

bobbym · 2011-05-24 06:02:13

Hi;

You are okay up to here.

The next step you made I do not follow at all.

Bob · 2011-05-24 06:38:18

I think it's an attempt to do

prior to de-logging both sides.

anonimnystefy : We know what log base you are using so leave it out to make your working clearer.

LATER EDIT:

This is ok. I made it -3/2<x<0 as well. LATER EDIT: NO IT IS'NT ... READ ON LATER POSTS!

One picky criticism which actually doesn't matter in practice.

square root(negative) is not possible in real numbers so your proof only works if

5x/2 + 4 ≥ 0 => x ≥ -8/5

This is ok for your answer so no problem but be careful in the future; you should check for this as it could have added a 'stricter' limit on x.

bobbym: with x = -5/4 I get 0.935414 > 0.75. LATER EDIT: NOW I THINK THIS IS WRONG!

Bob

Last edited by Bob (2011-05-24 21:48:25)

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Last Assessment Please Help Me!!

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