Post by: onur369 on February 03, 2011, 09:22:17 pm
So Ill just kick start it. I attached two questions below.
Post by: taiga on February 03, 2011, 09:32:26 pm
And if (3,3) is a stationary point you can also differentiate it and let it equal 0 for x=3 for another equation.
Post by: m@tty on February 03, 2011, 09:39:13 pm
a)
From inspection, b=-3 and c=3. Deduced by basic transformation rules. It's a basic cubic. And the point of inflection is given by (-b, c) in this case.
Sub, (2, 0):
b)
g(x) is a reflection in the y-axis, so
range is the same: [0,6]
Domain is [-4,-2]
c)
Height of 3.375 is the value of f(x), so find x such that f(x)=3.375
Now, the width is twice this distance as it is the value between f and g.
So the width is 7.067cm.
Post by: onur369 on February 03, 2011, 09:48:40 pm
Anyone know how to do the second one.
Post by: m@tty on February 03, 2011, 10:00:09 pm
Points given:
(2, 3) -
(5, 0) -
From (2)
Sub that into 1:
So,
As
To find the range you need to use calculus.
Let
So
So max occurs at t=10/3
So the range is
And to the closest 100m the max is 4600m, and this occurs at t=3.333 hours ie. 200 minutes
Post by: onur369 on February 03, 2011, 10:05:47 pm
Post by: m@tty on February 03, 2011, 10:22:41 pm
My only requirement is that you now get 50 :) (seriously, ask andiio)
Post by: onur369 on February 03, 2011, 10:25:58 pm
Post by: MrIraq on February 04, 2011, 12:11:58 am
hahahahaha, I would never get a 50. At most 45 if I study 24/7. But a 40 will be nice.
dream on...
Post by: onur369 on February 04, 2011, 10:04:27 pm
hahahahaha, I would never get a 50. At most 45 if I study 24/7. But a 40 will be nice.
dream on...
Stfu retard, go get a 35 in legal without knowing what a plaintiff and verdict is..
Post by: onur369 on February 06, 2011, 03:47:58 pm
If the functions (http://www4b.wolframalpha.com/Calculate/MSP/MSP419219e701df0d1hbf99000039999gb6ai52ib32?MSPStoreType=image/gif&s=51&w=156&h=20) and (http://www4b.wolframalpha.com/Calculate/MSP/MSP130519e704abb7bg3f5d0000375358ae4b8db0a8?MSPStoreType=image/gif&s=47&w=290&h=20) both cross the x-axis at -1, determine the values for a and b.
NOTE: First function equals to f(x)
Post by: Greatness on February 06, 2011, 03:51:36 pm
Hey guys, Just got caught up with a question. I can do it but Im wondering if there is a quicker way to do it.SUb x=-1 and y=0 into f(x) and g(x) then solve for a and b via simutaneous equations.
If the functions (http://www4b.wolframalpha.com/Calculate/MSP/MSP419219e701df0d1hbf99000039999gb6ai52ib32?MSPStoreType=image/gif&s=51&w=156&h=20) and (http://www4b.wolframalpha.com/Calculate/MSP/MSP130519e704abb7bg3f5d0000375358ae4b8db0a8?MSPStoreType=image/gif&s=47&w=290&h=20) both cross the x-axis at -1, determine the values for a and b.
NOTE: First function equals to f(x)
Post by: onur369 on February 06, 2011, 04:09:32 pm
Post by: Greatness on February 06, 2011, 04:25:05 pm
The results I obtained were: a= -5 and b= -7. Correct?I just chucked it all into my calculator, i got a=4 and b=-16. Maybe my method is wrong? :s
Post by: xZero on February 06, 2011, 04:34:55 pm
The results I obtained were: a= -5 and b= -7. Correct?I just chucked it all into my calculator, i got a=4 and b=-16. Maybe my method is wrong? :s
your method should be correct and i got a=4 and b=-16 as well
Post by: onur369 on February 06, 2011, 04:57:46 pm
(-1)^4 +a(-1)^3 - (-1)^2 + b(-1) -12=0
1-a-1-b-12=0
-a-b-12=0
a=-b-12
g(-1)=0
(-1)^4 (-b-12)(-1)^3 -9(-1) +b(-1)^2 +29(-1)+3- =0
1+b+12+9+b-29+30=0
14+2b=0
2b=-14
b=-7
sub in b into a
a= -(-7)-12
a= 7-12
a=-5.
Thats what I did.
Post by: Greatness on February 06, 2011, 05:08:13 pm
g(-1)= 1+-(a-9) + (b+9) -29=30=0 => -a+b=-20 {2}
f: a=-b-12 rearragne {1} sub into {2}
-(-b-12)+b=-20 => 2b+12=-20 => 2b=-32 => b=-16 sub into {1}
a=-(-16)-12=4
a=4 and b=-16
Post by: luken93 on February 06, 2011, 05:15:44 pm
-------
Post by: onur369 on February 06, 2011, 05:43:20 pm
Also swarley might want to double check you calculations, especially towards the end of your working out. Seems wrong to me. Since when does -b +b = -2b :/
-(b+12)+b=20 => -2b=32 => b=-16 sub into {1}
Post by: Greatness on February 06, 2011, 06:07:33 pm
Cheers.haha yeah bad working there :x fixed.
Also swarley might want to double check you calculations, especially towards the end of your working out. Seems wrong to me. Since when does -b +b = -2b :/
-(b+12)+b=20 => -2b=32 => b=-16 sub into {1}
Post by: Andiio on February 06, 2011, 07:17:26 pm
Ah it's cool.. It's good to help everyone out. :)
My only requirement is that you now get 50 :) (seriously, ask andiio)
LOL OMG m@tty you still remember that? :P
Post by: onur369 on February 06, 2011, 07:19:35 pm
Post by: Halil on February 06, 2011, 08:06:20 pm
Post by: onur369 on February 06, 2011, 09:43:02 pm
Post by: MrIraq on February 06, 2011, 09:51:09 pm
Onur, study together, find solutions together! -_-.
Of course bro. We need 94+
nerds..
Post by: onur369 on February 06, 2011, 09:59:30 pm
Post by: m@tty on February 06, 2011, 10:41:07 pm
Ah it's cool.. It's good to help everyone out. :)
My only requirement is that you now get 50 :) (seriously, ask andiio)
LOL OMG m@tty you still remember that? :P
You didn't think you were going to get out of it that easily did you.
Btw, your aims equate to a 99.90 :P haha (aggregate of 207.2)
But with the 50 for Spesh you break the 99.95 (209.2)
Post by: Andiio on February 06, 2011, 10:50:03 pm
Ah it's cool.. It's good to help everyone out. :)
My only requirement is that you now get 50 :) (seriously, ask andiio)
LOL OMG m@tty you still remember that? :P
You didn't think you were going to get out of it that easily did you.
Btw, your aims equate to a 99.90 :P haha (aggregate of 207.2)
But with the 50 for Spesh you break the 99.95 (209.2)
You atar-calc'ed me? LOL
I only aspire to get into my course though, nothing else :)
Post by: onur369 on February 07, 2011, 08:54:22 pm
Ok here is one ridiculously easy question, I know how to do, but I dont want to spend more than 3-4minutes doing.
Expand: (2y-3x)^5
Any quicker ways, rather than doing it the long way and wasting valuable test time.
Post by: xZero on February 07, 2011, 09:01:26 pm
Post by: onur369 on February 07, 2011, 09:04:29 pm
Post by: pi on February 07, 2011, 09:18:05 pm
Post by: onur369 on February 07, 2011, 09:20:30 pm
Post by: brightsky on February 07, 2011, 09:28:48 pm
Post by: pi on February 07, 2011, 09:39:40 pm
If you use the binomial theorem enough, it won't take long, definitely within the time you specified. Also just a tip, make use of Pascal's triangle. It can cut down the time by heaps.
True, just write down Pascal's triangle (shouldn't take more than 30 seconds for up to degree 5)
Post by: m@tty on February 07, 2011, 09:47:31 pm
Always go pascals though it's a lot faster as long as it's less than degree 10 or so..
Post by: onur369 on February 16, 2011, 08:44:22 pm
Post by: pi on February 16, 2011, 08:59:35 pm
By finding the red area, they just mean the two lines on the outside (not the actual area), from memory, y=|3x/2|+6 where x = [-2,2] for the first part. Just 'move this around' for part b.
Post by: onur369 on February 16, 2011, 09:19:53 pm
Post by: onur369 on March 09, 2011, 07:03:50 pm
If y=g(x) for x ∈ [-1,3] and y=h(x) for x ∈ [0,4] , then y=g(x) + h(x) is defined for .... ?
Post by: Water on March 09, 2011, 07:06:50 pm
Ohh, yeah, I forgot to put working out :)
y= g(x) Domain = [-1,3]
y= h(x) Domain = [0,4]
To find the intersection would be quite simple
You look at all the x values that g(x) can have, that is, -1,0,1,2,3
You look at all the x values that h(x) can have, that is, 0,1,2,3,4
Then you do look and match,
We know that g(x) and h(x) have the same domains as 0,1,2,3
Hence we know that the intersection of g(x) and h(x) for it to be defined is 0,1,2,3 = [0,3]
Easier method, is to put the domain on a number line, and see which domains match each other :)
Post by: Halil on March 09, 2011, 07:15:54 pm
Post by: pi on March 09, 2011, 07:17:54 pm
The intersection of the two domains.
A handy trick to know that. Works for sum, difference and product functions.
Post by: onur369 on March 09, 2011, 07:18:47 pm
Post by: Pandabear on March 11, 2011, 04:47:33 pm
Could use some help on this one.
Post by: Halil on March 11, 2011, 04:53:19 pm
Post by: onur369 on March 11, 2011, 04:56:08 pm
Than substitue -3 into x into the intial equation and make it equal to 144, solve it for b.
Once you solve it for B, add in the a value you found from the first process and substitue it into the second one to find b.
After finding b you can just sub it into the equation of a= .....
Then you will find a,b
a=2 b=-5
Post by: Pandabear on March 11, 2011, 05:00:05 pm
Post by: Abdulhai on March 11, 2011, 05:55:30 pm
Post by: onur369 on March 11, 2011, 06:00:06 pm
Post by: Pandabear on March 11, 2011, 06:02:51 pm
Post by: pi on March 11, 2011, 06:05:08 pm
Post by: luken93 on March 11, 2011, 06:28:15 pm
We know that
Sub a=3 into (1):
Sub a=3 and b=10 into 2 to check:
Post by: onur369 on March 11, 2011, 06:52:08 pm
The equation of the image of y=x^2 under a shift to the left of 4 units, a dilation of by a factor of 2 in the y-direction and a shift downwards of 3 units is:
2y=(x+4)^2 -3
or
y=2(x+4)^2 -3
Post by: luken93 on March 11, 2011, 06:57:17 pm
Post by: onur369 on March 11, 2011, 06:58:32 pm
Post by: Pandabear on March 11, 2011, 07:07:54 pm
Let
We know that
Sub a=3 into (1):
Sub a=3 and b=10 into 2 to check:
That was an awesome response thanks.
Post by: Pandabear on March 11, 2011, 07:10:39 pm
Got caught up with a stupid question, probably my most weakest area.
The equation of the image of y=x^2 under a shift to the left of 4 units, a dilation of by a factor of 2 in the y-direction and a shift downwards of 3 units is:
2y=(x+4)^2 -3
or
y=2(x+4)^2 -3
answer should be 2y=(x+4)^2 -3
Post by: onur369 on March 11, 2011, 07:12:37 pm
Post by: onur369 on March 11, 2011, 08:13:39 pm
(log3(x))^2 -4log3(x) +3 =0
Post by: luffy on March 11, 2011, 08:13:54 pm
Got caught up with a stupid question, probably my most weakest area.
The equation of the image of y=x^2 under a shift to the left of 4 units, a dilation of by a factor of 2 in the y-direction and a shift downwards of 3 units is:
2y=(x+4)^2 -3
or
y=2(x+4)^2 -3
Actually, the answer is:
--> A dilation "in the y-direction" is the same as a dilation "from the x-axis".
Post by: luffy on March 11, 2011, 08:21:01 pm
Post by: onur369 on March 11, 2011, 08:23:17 pm
Post by: Halil on March 11, 2011, 08:36:47 pm
Post by: luken93 on March 11, 2011, 08:50:35 pm
(x,y)
(x-4, y) a shift to the left of 4 units
(x-4, 2y) a dilation of by a factor of 2 in the y-direction
(x-4, 2y-3) a shift downwards of 3 units is:
x' = x-4 => x = x'+4
y' = 2y-3 => y=(y'+3)/2
(y + 3)/2 = (x - 4)^2
y - 3 = 2(x - 4)^2
y = 2(x - 4)^2 + 3
EDIT: Woops, didnt even see luffy's response
Post by: Pandabear on March 11, 2011, 09:05:16 pm
is the answer x=-3 , x=2
or x=3 ,x=2
Post by: m@tty on March 11, 2011, 09:11:18 pm
ie.
However, if you mean -2^x=-8, then x=3
Post by: luffy on March 11, 2011, 09:24:28 pm
2^x=-8 and 2^x=4Think about your question logically. Sub in any value of x, for 2^x. For example, if x = 2,
is the answer x=-3 , x=2
or x=3 ,x=2
If you sub in x = 3
If you sub in x = -3,
you will get
Now, if you look at the values of x that i subbed in, did you see any negative numbers? No, there were no negative answers. Therefore, for 2^x = -8, there are no numbers of x that will make 2^x = a negative number. Thus, for 2^x = -8, x has no solutions.
For 2^x = 4, you could use logarithms; however, its quite trivial to see that 2^2 = 4, meaning that x = 2.
Post by: onur369 on March 15, 2011, 10:10:14 pm
I attached it below.
Post by: Respect on March 15, 2011, 10:19:12 pm
Hey guys, got caught up with another question.
I attached it below.
i bet 10 buxs the answer is 2x, lol
Post by: nacho on March 15, 2011, 10:24:07 pm
Post by: luken93 on March 15, 2011, 10:24:28 pm
Long story short, if you can see the trends it leads you to 2x, because 2(x + y)/2 = x + y
Similarly, (2x + 2y)/2 = x + y
Post by: onur369 on March 15, 2011, 10:25:01 pm
Hey guys, got caught up with another question.
I attached it below.
i bet 10 buxs the answer is 2x, lol
No shit sherlock if you check BOB.
Post by: Respect on March 15, 2011, 10:25:10 pm
Yea, D.
which lanuage is this ??
Quote
"No! I must kill the demons" he shouted.
The radio crackered "No, John. You are the demons"
And then John was a zombie.
Post by: Respect on March 15, 2011, 10:29:38 pm
Hey guys, got caught up with another question.
I attached it below.
i bet 10 buxs the answer is 2x, lol
No shit sherlock if you check BOB.
nah i checked abbas actually. lol :p
hey idk how to solve it either, wth is it./..
Post by: luffy on March 15, 2011, 11:07:30 pm
Hey guys, got caught up with another question.
I attached it below.
i bet 10 buxs the answer is 2x, lol
No shit sherlock if you check BOB.
nah i checked abbas actually. lol :p
hey idk how to solve it either, wth is it./..
This question is simply trial and error. Just try all the options and see if it satisfies the equation:
a)
As you can see, this does not satisfy the equation.
However, if you try the same process for option D, the two equations will be equal
Post by: brightsky on March 15, 2011, 11:12:01 pm
Post by: onur369 on March 15, 2011, 11:16:56 pm
5 + log2 (5x) = log2 (y)
ffs i have problems converting 5 into a log2 form
Post by: brightsky on March 15, 2011, 11:22:49 pm
log2(2^5) + log2(5x) = log2(y)
log2(2^5*5x) = log2(y)
y = 2^5*5x = 140x
Post by: onur369 on March 15, 2011, 11:26:16 pm
Post by: nacho on March 16, 2011, 02:12:31 pm
I need help with:I'm not bashing on you, but i suggest that you have another look at how the log works, it is quite crucial for you to know these first-level basics if you want to go on and do harder problems.
5 + log2 (5x) = log2 (y)
ffs i have problems converting 5 into a log2 form
Post by: Mao on March 17, 2011, 11:31:21 pm
C'mon guys, this is the maths board for crying out loud.
Please keep these confrontations in pm, you guys (and ladies) are mature enough to be able to sort these things out without making a racket. I'm sure a small misunderstanding such as this can be settled by a simple message exchange. There's no need to publicize it and give it the 'wolf pack' feeling.
If you would like to voice without a direct confrontation, or feel a pm is too direct, please do not hesitate to contact any of the mods and we will help to mediate.
Back on topic please. =]
Post by: Respect on March 17, 2011, 11:40:50 pm
Moderator Action: off-topic posts have been deleted, original advice by nacho is preserved. Everyone whose post(s) have been deleted, consider this a serious reminder.
C'mon guys, this is the maths board for crying out loud.
Please keep these confrontations in pm, you guys (and ladies) are mature enough to be able to sort these things out without making a racket. I'm sure a small misunderstanding such as this can be settled by a simple message exchange. There's no need to publicize it and give it the 'wolf pack' feeling.
If you would like to voice without a direct confrontation, or feel a pm is too direct, please do not hesitate to contact any of the mods and we will help to mediate.
relax, mao... we're just messing, nacho is our freind, dont u like to tease ur freinds sometimes :)
rite nacho?
Post by: Mao on March 17, 2011, 11:57:01 pm
Moderator Action: off-topic posts have been deleted, original advice by nacho is preserved. Everyone whose post(s) have been deleted, consider this a serious reminder.
C'mon guys, this is the maths board for crying out loud.
Please keep these confrontations in pm, you guys (and ladies) are mature enough to be able to sort these things out without making a racket. I'm sure a small misunderstanding such as this can be settled by a simple message exchange. There's no need to publicize it and give it the 'wolf pack' feeling.
If you would like to voice without a direct confrontation, or feel a pm is too direct, please do not hesitate to contact any of the mods and we will help to mediate.
relax, mao... we're just messing, nacho is our freind, dont u like to tease ur freinds sometimes :)
rite nacho?
As I said, please keep it on topic.
Post by: kefoo on March 19, 2011, 01:50:39 pm
edit: didnt see it intersected at (0,0)
another question though:
Prove that if logr(P) = q and logq(r) = p then logq(p) = pq
Post by: pi on March 19, 2011, 02:41:47 pm
There's probably an easier way, but meh, this one works
Post by: kefoo on March 19, 2011, 03:12:29 pm
nice! thanks alot
There's probably an easier way, but meh, this one works
just one more if thats ok
y = ae^x + b with points (1,14) and (0,0)
i did simultaneous eqns but i get the wrong answer =.=
Post by: pi on March 19, 2011, 03:20:31 pm
Hopefully thats right
Post by: kefoo on March 20, 2011, 11:36:53 am
ohh thats how you do it thanks.
Hopefully thats right
weekend is probably getting to me =.=
Post by: kefoo on March 22, 2011, 06:38:10 pm
if f(x) = 1 - e^(-x)
find the inverse.
i swap the x and y as usual, but dunno how to get the log and stuff..
Post by: Water on March 22, 2011, 06:43:13 pm
x = 1 - e^(-y)
x - 1 = -e^(-y)
1-x = e^(-y)
loge(1-x) = -y
-loge(1-x) = y
Post by: onur369 on March 22, 2011, 06:43:54 pm
y=1-1/e^x <- Make all indices positive before any other steps.
Substiute x and y. Therefore, x=1-1/e^y
Move values around 1/e^y=1-x
Move again. e^y= 1/1-x
Loge both sides it will equal y= log_e(1/1-x)
f^-1(x)= log_e(1/1-x)
Post by: dooodyo on March 22, 2011, 11:06:14 pm
And yeah Water's solution is right.
Post by: m@tty on March 22, 2011, 11:12:45 pm
Post by: onur369 on March 24, 2011, 05:47:36 pm
Calculus question: Show that the derivative of y = k, where k is a constant, is zero.
What do we do and why do we do it?
Post by: Water on March 24, 2011, 05:50:17 pm
Given that k is a constant . such as 6 , 7, 8 ,9 ,10
The gradient = rise/ run
When its (0,6), the rise is still (1,6) which is 6
Therefore, there is 0 rise. Run is 1
Gradient: 0/1 = 0
Dy/dx = 0
Reasoning: There is no rise, however there is still run. when y = a constant
Post by: onur369 on March 24, 2011, 05:54:49 pm
Post by: kefoo on March 24, 2011, 06:27:03 pm
1.Find the values of a and b such that the graph of y = ae^(bx) goes through (3,10) and (6,50)
2.Find the values of a and b such that the graph of y = alog2(x+b) goes through the points (8,10) and (32,14)
Post by: Greatness on March 24, 2011, 06:37:01 pm
1) 10 = a*e^(3b) and 50 = a*e^(6b)
2) 10 = alog2(8+b) and 14 = alog2(32+b)
Post by: kefoo on March 24, 2011, 06:44:14 pm
Well you would sub those 2 coordinates into the equations which will give you two equations which you can solve simultaneously. You may be able to use the solve function on the CAS to make it quicker (This would only be if its tech active)yeah i can do it by CAS easily but im wondering where i stuffed up when doing it by hand xD
1) 10 = a*e^(3b) and 50 = a*e^(6b)
2) 10 = alog2(8+b) and 14 = alog2(32+b)
Post by: Greatness on March 24, 2011, 06:47:28 pm
Ill have a go at doing them
Post by: Mao on March 25, 2011, 01:17:12 am
Well you would sub those 2 coordinates into the equations which will give you two equations which you can solve simultaneously. You may be able to use the solve function on the CAS to make it quicker (This would only be if its tech active)
1) 10 = a*e^(3b) and 50 = a*e^(6b)
2) 10 = alog2(8+b) and 14 = alog2(32+b)
The first one can be solved algebraically,
10 = a e^(3b) ----- [1]
50 = a e^(6b) ----- [2]
[2]/[1]: 5 = e^(3b) --> b = Ln(5)/3, a = 2
And I have a sneaking suspicion q2 should be y = a log2(x) + b, where a=2 and b=4. Otherwise it must be solved on a calculator.
Post by: kefoo on March 26, 2011, 11:43:14 am
sin(x) = -0.45
dunno what to do because there arent any examples ._.
Post by: pi on March 26, 2011, 12:34:15 pm
Find all values of x between 0 and 2pi for which:
sin(x) = -0.45
dunno what to do because there arent any examples ._.
I think you'll have t use a calc to find x (and sin-1(9/20) has no surd value from my knowledge) and then use symmetry of the unit circle to find all the solutions.
Post by: onur369 on April 05, 2011, 05:56:13 pm
Post by: xZero on April 05, 2011, 06:00:52 pm
the gradient of PQ is
sub the points in and you'll get E
Post by: onur369 on April 05, 2011, 06:19:45 pm
Post by: xZero on April 05, 2011, 06:30:35 pm
The first step is to find
The gradient of the chord PQ is just asking you to find the change in y in respect of x from P to Q
Post by: onur369 on April 05, 2011, 07:44:10 pm
find derivative of:
4ln(sqroot 2-x^2)
Post by: xZero on April 05, 2011, 08:01:28 pm
use the chain rule, let
Edit: blah my bad
Post by: m@tty on April 05, 2011, 08:08:40 pm
because the original function is not defined when |x| is root 2, as it results in ln(0).
Post by: Respect on April 05, 2011, 09:00:46 pm
Find the derivative of that.. how do i used the product rule.. because i seem yo keep gettin the wrong answer which is//
3/(x-2)^2
but the right answer is E.. can someone solve it?
Post by: xZero on April 05, 2011, 09:05:00 pm
Post by: Respect on April 05, 2011, 09:07:18 pm
use the quotient rule instead :P
sorry i meant i was using the quotient rule... but i still seem to get it wrong.. idk what program ur using to write that math equations but what i do is, 1 sec. ill show u
Post by: jasoN- on April 05, 2011, 09:08:00 pm
or quotient rule
Post by: Respect on April 05, 2011, 09:16:06 pm
or quotient rule
ohh!! lol.. i was writing the forumla wrong.. instead i was writing
u`v - v`u
________
v^2
Post by: m@tty on April 05, 2011, 09:20:17 pm
use the quotient rule instead :P
Or product rule:
Post by: onur369 on April 05, 2011, 09:37:27 pm
Which method should I use for the question below
Post by: m@tty on April 05, 2011, 09:40:14 pm
As required.
Post by: onur369 on April 05, 2011, 09:41:12 pm
Post by: Respect on April 05, 2011, 09:57:06 pm
I would go
As required.
i got no idea how u did that ???
the second step.. how did u simplfy into 3(x+5)^2 ... .. .. .
im stuffed for the test tomrrow.
Post by: onur369 on April 05, 2011, 09:58:17 pm
Post by: jasoN- on April 05, 2011, 10:07:13 pm
I would go
As required.
i got no idea how u did that ???
the second step.. how did u simplfy into 3(x+5)^2 ... .. .. .
im stuffed for the test tomrrow.
Post by: xZero on April 05, 2011, 10:08:59 pm
i got no idea how u did that ???
the second step.. how did u simplfy into 3(x+5)^2 ... .. .. .
im stuffed for the test tomrrow.
use product rule,
take out the common factor
Post by: kefoo on April 09, 2011, 09:33:22 am
need help on this question::
1)a.Sketch graph of y = cos(x) and y = sqrt(3)sin(x) [Sketched] (0,2pi)
b. Find the coordinates of the points of intersection [Equated them to get tanx = 1/sqrt3, got pi/6 and 7pi/6 correct, just don't know how to get the y values :S
Post by: jasoN- on April 09, 2011, 09:44:56 am
Post by: kefoo on April 09, 2011, 12:47:51 pm
just sub pi/6 and 7pi/6 into either of the equations to get the y value (cos both graphs intersect at the same coordinate)but wouldn't sin(pi/6) be different to cos(pi/6)?
Post by: luffy on April 09, 2011, 12:54:19 pm
just sub pi/6 and 7pi/6 into either of the equations to get the y value (cos both graphs intersect at the same coordinate)but wouldn't sin(pi/6) be different to cos(pi/6)?
Yes. However,
The equation you had was not
Hope I helped.
Post by: kefoo on April 09, 2011, 12:59:34 pm
just sub pi/6 and 7pi/6 into either of the equations to get the y value (cos both graphs intersect at the same coordinate)but wouldn't sin(pi/6) be different to cos(pi/6)?
Yes. However,
The equation you had was not. It was
Hope I helped.
oh right, forgot about sqrt3... x_x careless mistakes.
Got another q..
tan(2x-pi/4) = sqrt(3) solve for 0< x < 2pi
Post by: luffy on April 09, 2011, 01:14:32 pm
just sub pi/6 and 7pi/6 into either of the equations to get the y value (cos both graphs intersect at the same coordinate)but wouldn't sin(pi/6) be different to cos(pi/6)?
Yes. However,
The equation you had was not
Hope I helped.
oh right, forgot about sqrt3... x_x careless mistakes.
Got another q..
tan(2x-pi/4) = sqrt(3) solve for 0< x < 2pi
Below is how I would do it:
The same principle applies to this question, however, you want all the solutions of
Tell me if I didn't explain anything clear enough (sorry if I made any errors).
Hope I helped.
Post by: panicatthelunchbar on April 09, 2011, 04:33:52 pm
11) A patient suffering with hypothermia after being lost in the snow for 2 days is covered in
thermal blankets and warmed so that his body temperature can return to normal levels.The change in body temperature (T°C) from the initial temperature that occurs during thermal treatment is modelled by the equation T = A loge (t − b) where t represents the time in minutes. The initial body temperature is 35.7°C and it reaches 36.1°C ten minutes after treatment is initiated.
(a) What is the value ofT when t = 0?
(b) Find the value of b.
(c) Find the value of A.
(d) What will be the body temperature after 20 minutes?
(e) How long will it take for the body temperature to reach 36.8°C? Write your answer
in hours and minutes.
thanks :)
Post by: onur369 on April 14, 2011, 06:25:22 pm
Question 6:
A cylinder is expanding in such a way that is length h is always double the radius r of its
base. Find the rate of change of its volume V with respect to:
a) r.
b) h.
Post by: brightsky on April 14, 2011, 06:35:11 pm
V = pi*r^2 h
But h = 2r (which means dh/dr = 2), r = h/2
So V = pi*(h/2)^2 h = pi/4*h^3 (which means dV/dh = 3pi/4*h^2)
So dV/dr = 3pi/4 * h^2*2 = 3pi/2*h^2
b) dV/dh = dV/dr * dr/dh
V = pi*r^2 h = pi*r^2(2r) = 2pi*r^3, dV/dr = 6pi*r^2
r = h/2, dr/dh = 1/2
So dV/dh = 1/2*6pi*r^2 = 3pi*r^2
Post by: onur369 on April 14, 2011, 06:35:41 pm
Post by: kefoo on April 14, 2011, 07:22:03 pm
tan (3x - pi/6) = -1
solve for x..
i get the wrong answer :S
Post by: Souljette_93 on April 14, 2011, 07:40:56 pm
having some trouble here..
tan (3x - pi/6) = -1
solve for x..
i get the wrong answer :S
My methods is a bit rusty but I'll give it a go.
So first do tan^-1 on both sides:
3x-pi/6=-pi/4
Add pi/6 to both sides:
3x=-pi/12
Divide by three:
X=-pi/36 +npi
That is a general equation, since you did not provide a
Particular domain
Post by: brightsky on April 14, 2011, 08:10:11 pm
tan(3x - pi/6) = -1
Let 3x - pi/6 = u
tan(u) = -1
u = pi*n-pi/4, where n E Z
so 3x - pi/6 = pi*n - pi/4
3x = pi*n - pi/12 = (12pi*n - pi)/12
x = (12pi*n - pi)/36
Post by: onur369 on April 25, 2011, 11:47:41 am
A wine glass is being Þ lled with wine at a rate of 8 cm3/s. The volume, V cm3, of wine in the glass when the depth of wine in the glass is x cm is given by V = 4x 32. Find the rate at which the depth of wine in the glass isincreasing when the depth is 4 cm.
Any quick method for it? Also only 42% got this question correct
Post by: luffy on April 25, 2011, 12:43:18 pm
- That is a good thing, as long as you know the course properly.
This is an application of chain rule question:
At x = 4,
Therefore, the depth is increasing at a rate of
Hope I helped.
Post by: onur369 on April 25, 2011, 12:50:34 pm
Post by: onur369 on April 26, 2011, 12:35:48 pm
Post by: xZero on April 26, 2011, 01:09:57 pm
Post by: luffy on April 26, 2011, 01:39:18 pm
You gotta use chain rule now (I just did it in one step.... sub
Note: You can also use product rule from the original equation, but I think that would be a bit longer.
For maximums,
You can use your sign table, but obviously r = 0 will not produce a maximum volume and the negative r value is rejected lol.
Therefore for Max Volume,
Sub this r-value back into the volume equation:
At
I hope I didn't make an error somewhere at the start.. lol
Post by: onur369 on April 26, 2011, 01:57:47 pm
Post by: hello_kitty on April 27, 2011, 08:22:01 pm
i.e.y = -1/x + 2
I know the y intercept of the hyperbola is (0,-1/2)
but when it comes to figuring out the x intercept it is 'false'
so what would i put under X int?? Would i just use x = 1 or another point?
Thanks :D
Post by: onur369 on April 27, 2011, 08:29:08 pm
find x int y=0
0 =-1/x +2
-2=-1/x
swap x
and -1/-2 thus being 1/2.
When trying to find the y int we sub in 0 but -1/0 is underfined and since there is no such thing as undefined + 2 it has no y int, its asympote is y=0
Post by: brightsky on April 28, 2011, 05:55:53 pm
Post by: Respect on April 28, 2011, 07:21:20 pm
if sin(x) = 0.3, cos(a)=0.5, tan(b) = 2.4 and x,a and b are in the first quadrant, find the value of the following.
a.) cos(π-x)
b.) sin(π-a)
π is pi
how u can solve these im not know?
Post by: brightsky on April 28, 2011, 07:39:45 pm
sin(pi - a) = sin(a) from symmetry. using the above identity, sin(a) = sqrt(1 - cos^2(a)) = sqrt(1 - 0.5) = sqrt(0.5)
EDIT: see below
Post by: jbebbo on April 28, 2011, 07:44:29 pm
cos(pi - x) = -cos(x) by symmetry. we know that sin^2(x) + cos^2(x) = 1, so cos(x) = sqrt(1 - sin^2(x)) = sqrt(1 - 0.3) = sqrt(0.7).
sin(pi - a) = sin(a) from symmetry. using the above identity, sin(a) = sqrt(1 - cos^2(a)) = sqrt(1 - 0.5) = sqrt(0.5)
Don't you mean
cos(x) = sqrt(1 - sin^2(x)) = sqrt(1 - 0.09) = sqrt(0.91) therefore cos(pi-x)=-sqrt(0.91)?
similarly
sin(a) = sqrt(1 - cos^2(a)) = sqrt(1 - 0.25) = sqrt(0.75)?
Post by: brightsky on April 28, 2011, 07:53:00 pm
cos(pi - x) = -cos(x) by symmetry. we know that sin^2(x) + cos^2(x) = 1, so cos(x) = sqrt(1 - sin^2(x)) = sqrt(1 - 0.3) = sqrt(0.7).
sin(pi - a) = sin(a) from symmetry. using the above identity, sin(a) = sqrt(1 - cos^2(a)) = sqrt(1 - 0.5) = sqrt(0.5)
Don't you mean
cos(x) = sqrt(1 - sin^2(x)) = sqrt(1 - 0.09) = sqrt(0.91) therefore cos(pi-x)=-sqrt(0.91)?
similarly
sin(a) = sqrt(1 - cos^2(a)) = sqrt(1 - 0.25) = sqrt(0.75)?
oh lol..haha yes. thanks jbebbo!
Post by: Respect on April 28, 2011, 08:27:57 pm
cos(pi - x) = -cos(x) by symmetry. we know that sin^2(x) + cos^2(x) = 1, so cos(x) = sqrt(1 - sin^2(x)) = sqrt(1 - 0.3) = sqrt(0.7).
sin(pi - a) = sin(a) from symmetry. using the above identity, sin(a) = sqrt(1 - cos^2(a)) = sqrt(1 - 0.5) = sqrt(0.5)
Don't you mean
cos(x) = sqrt(1 - sin^2(x)) = sqrt(1 - 0.09) = sqrt(0.91) therefore cos(pi-x)=-sqrt(0.91)?
similarly
sin(a) = sqrt(1 - cos^2(a)) = sqrt(1 - 0.25) = sqrt(0.75)?
uhh im sorry bro, i still havent understood.. where did u get the formula cos(x) = sqrt(1 - sin^2(x))? and sin(a) = sqrt(1 - cos^2(a)) ... i dont have anything like that in my book, nor do they explain how to do these questions...
Post by: xZero on April 28, 2011, 08:33:08 pm
cos(x)^2 + sin(x)^2 = 1, sin(x)^2 = 1 - cos(x)^2, sin(x) = sqrt(1 - cos^2(x))
Post by: panicatthelunchbar on April 28, 2011, 09:52:09 pm
If 3+log2(4x)=log2 y, find y in terms of x
Post by: brightsky on April 28, 2011, 09:59:54 pm
log2(2^3) + log2(4x) = log2 y
log2(8*4x) = log2 y
y = 32x
Post by: panicatthelunchbar on April 28, 2011, 11:20:36 pm
Post by: onur369 on May 10, 2011, 07:45:49 pm
The -1 is confusing me.
Post by: david10d on May 10, 2011, 07:48:31 pm
Post by: onur369 on May 10, 2011, 08:51:09 pm
Post by: david10d on May 10, 2011, 08:55:21 pm
It's just expressed in degrees instead of radians.
Post by: onur369 on May 10, 2011, 09:46:58 pm
Post by: luken93 on May 10, 2011, 09:57:46 pm
Thinking about the graph of sin(x), it has a max at (pi/2, 1) and min at (3pi/2, -1)
Thus, (b - 1)/2 =1
b - 1 = 2
b = 3
Post by: Respect on May 10, 2011, 10:09:07 pm
sin(x) = (b - 1)/2^
Thinking about the graph of sin(x), it has a max at (pi/2, 1) and min at (3pi/2, -1)
Thus, (b - 1)/2 =1
b - 1 = 2
b = 3
whaa? didnt understand that :/
do u know how to graph y=2sin( l x-pi/2 l ) to one period?
Post by: luken93 on May 10, 2011, 10:17:25 pm
I rearranged the equation to make it sin(x) = ....sin(x) = (b - 1)/2^
Thinking about the graph of sin(x), it has a max at (pi/2, 1) and min at (3pi/2, -1)
Thus, (b - 1)/2 =1
b - 1 = 2
b = 3
whaa? didnt understand that :/
do u know how to graph y=2sin( l x-pi/2 l ) to one period?
Now think about the graph of sin(x) from [0, 2pi]
--
/ \
/ \ .[2pi, 0]
\ /
\ /
--
For it to have one solution only, it must hit the graph once only. The only points at which this occurs are at the min and max of the graph, ie at [pi/2, 1] and [3pi/2, -1]
Since b is a positive number, then it must be the point pi/2, 1
Solving for this yields b = 3
---------
As for graphing that, draw an imaginary line at x = pi/2
Then graph the function as normal to the right of this line.
Because it is modulus, it will be reflected on this line (think about when x = 0 for example).
Post by: Respect on May 10, 2011, 10:28:28 pm
Post by: onur369 on May 10, 2011, 10:40:19 pm
Post by: brightsky on May 10, 2011, 10:45:37 pm
its an absolute function, thats not right.
i believe just sketch the normal function f(x) = 2sin(x), find f|x| = 2sin|x| --> just wipe out everything to the left of the y-axis and reflect the stuff on the right, and then translate across to the right pi/2 units.
or more generally
2sin|x - pi/2|
= 2 sin(x - pi/2), x - pi/2 >= 0, x >= pi/2
= 2 sin(pi/2 - x), x - pi/2 < 0, x < pi/2
sketch from there
Post by: luken93 on May 10, 2011, 10:53:36 pm
its an absolute function, thats not right.The -------- seperates yours and onur's q?
Post by: Respect on May 12, 2011, 12:42:21 am
His talking about my question -.-
he?
Post by: onur369 on May 14, 2011, 07:51:20 pm
I attached the initial question but its following questions are as followed:
a) Find the area, A, of rectangle XYZW in terms of a.
b) Find the maximum value of A and the value of a for which this occurs.
Post by: xZero on May 14, 2011, 07:59:49 pm
a) To find the area, we need the length and the width. The length of the retangle is from the point X(-a,0) to W(a,0) so if we find the difference in their x-value, it will give us the length. Hence the length is a-(-a)=2a. The width is the difference in y-value from X(-a,0) to Y(-a,b) which is b. However we want to express the area in terms of a so we must find a relationship between a and b. Note the equation y=9-3x^2, where y=b and x=a. Thus b=9-3a^2. Area = LxW, Area=2a x (9-3a^2), Area=18a-6a^3
b)Now diff the area, find the maximum value for a
Post by: onur369 on May 19, 2011, 07:32:21 pm
Post by: onur369 on May 19, 2011, 08:03:57 pm
Post by: gossamer on May 19, 2011, 08:18:18 pm
For there to be a real solution for the equation, sqrt(1-x) must be smaller or equal to 5 ( i.e. (-inf,5] ). But the sqrt can't be equal to a real negative number so it must be equal to [0,5]
You should be able to get the answer from there :)
Post by: luken93 on May 19, 2011, 08:24:23 pm
For it to work, let
Implied domain would then be
For it to exist, domain of u must be [-inf, 5]
Hence the max is 5
Hence, the domain is [-24, 1]
Post by: onur369 on May 19, 2011, 08:38:19 pm
I have another question:
If g: [k,4] -> R, g(x) = -(x-2)^2 +3
a) Find the smallest value for k such that g^-1 exists
b) Find the rule that defines g^-1(x), giving the domain.
Post by: brightsky on May 19, 2011, 08:40:36 pm
1-x => 0, x <= 1
5-sqrt(1-x) >= 0, sqrt(1-x) =< 5, 1-x =< 25 (note this is justified because both sides of the inequation are positive), x => -24
Hence we have -24 =< x =< 1
notation: => is greater or equal to, etc.
Post by: brightsky on May 19, 2011, 08:41:23 pm
(the answer you get should be 2.)
as for the rule, just swap the "y" and "x" around then make the domain [-1, 3] --> the range of the original function g(x)
Post by: onur369 on May 20, 2011, 11:47:34 am
Find exact solutions of equation:
sin(3x)-cos(3x)=0 for 0 less than equal to x less than equal to pi
Post by: kamil9876 on May 20, 2011, 02:18:53 pm
Post by: onur369 on June 21, 2011, 09:27:56 pm
VCAA 2007:
P is the point on the line 2x+y-10=0 such that the length of OP, the line segment from the origin O to P is a minimum. Find the coordinates of P and this minimum length.
VCAA 2006:
A normal to the graph of y=x^.5 has the equation y=-4x+a, where a is a real constant. Find the value of a.
Post by: luken93 on June 21, 2011, 09:36:39 pm
The shortest distance will always occur when the gradient is a normal to the line in question
ie, gradient = 1/2
Since you know it goes thru the origin, the equation is y = 1/2x
Finally, find point, the point in which they intersect;
1/2x = -2x + 10
5/2x = 10
5x = 20
x = 4
(4, 2)
The length is sqrt{ (y2- y1)^2 + (x2 - x1)^2)
= sqrt(4 + 16)
= sqrt(20)
= 2sqrt(5)
Post by: luken93 on June 21, 2011, 09:39:34 pm
y = sqrt(x)
y' = 1/(2sqsrt(x))
Find when y' = 1/4
x = 4
Hence, y = -4x + a passes thru (4, 2)
a = 18
Hopefully I haven't made any mistakes :P
Post by: onur369 on July 03, 2011, 01:43:53 pm
In exam 2 type questions, multiple choice there are questions like.
i.e: where p and q are real constants, have a unique solution provided or when no solution.
What do I do for each ? Like when it asks for a unique solution or no solution, Im assuming we use the discriminant?
Post by: xZero on July 03, 2011, 01:53:51 pm
Post by: onur369 on July 19, 2011, 05:03:20 pm
Question 1: Find f(x), if f'(x)=5sin(2x) and f(pi)=-1
- I do the first section correct, I integrate it and get -2.5cos(2x) its just the f(pi)=-1 part of the question
Question 2: A curve has the gradient dy/dx=ksin(3x)-3 where k is a constant. and a stationary point (pi/2,-2)
Find: a) the value of k, I attempted this I got -3, I am not sure if that is correct
b) Find the equation of the curve.
Post by: xZero on July 19, 2011, 05:13:15 pm
2.a) correct b)anti diff it, sub in (pi/2,-2) to find c
Post by: onur369 on July 19, 2011, 05:22:55 pm
Post by: b^3 on July 19, 2011, 05:56:25 pm
sub in
Post by: onur369 on July 19, 2011, 05:59:49 pm
Post by: onur369 on July 20, 2011, 08:31:44 pm
I am not good with memorising the exact values and now it has come across in integration. How do you remember the values?
Post by: tony3272 on July 20, 2011, 09:13:09 pm
Post by: david10d on July 20, 2011, 09:27:00 pm
Post by: pi on July 20, 2011, 09:30:40 pm
Look at the exact values table and you can see the pattern of square roots ascending (sin) or descending (cos).
That's what I do, and then sin/cos for tan (or memorise)
Post by: onur369 on July 20, 2011, 09:35:28 pm
(http://i.imgur.com/MT3S5.jpg)
Post by: pi on July 20, 2011, 09:38:29 pm
I'm stuck with an integral question, only problem here is how to deal with the denominator :/
(http://i.imgur.com/MT3S5.jpg)
Remember the ax+b special rules? Try looking those up :)
(don't have text on me, and I forget :P )
Post by: onur369 on July 20, 2011, 09:39:59 pm
Post by: onur369 on July 20, 2011, 10:08:13 pm
a) x=0 to x=1 , I did this and I got 2/3, not 100% sure.
What I am stuck on is the next two
b) x=1 to x=2
c) x=0 to x=2
^ I am not exactly sure what to do ? Do I draw a new line ?
Post by: tony3272 on July 20, 2011, 10:16:47 pm
Sketch the graph of y=1-x^2, showing all the intercepts. Find the area between the curve and the x axis fromYour answer is correct for part a. For part b), you just do the same method as part a, but just set your terminals as 1 and 2 rather than 0 and 1.
a) x=0 to x=1 , I did this and I got 2/3, not 100% sure.
What I am stuck on is the next two
b) x=1 to x=2
c) x=0 to x=2
^ I am not exactly sure what to do ? Do I draw a new line ?
Post by: enwiabe on July 21, 2011, 05:22:07 am
Sketch the graph of y=1-x^2, showing all the intercepts. Find the area between the curve and the x axis fromYour answer is correct for part a. For part b), you just do the same method as part a, but just set your terminals as 1 and 2 rather than 0 and 1.
a) x=0 to x=1 , I did this and I got 2/3, not 100% sure.
What I am stuck on is the next two
b) x=1 to x=2
c) x=0 to x=2
^ I am not exactly sure what to do ? Do I draw a new line ?
Ah, but it wants the area, not a simple evaluation of the integral. Think about where an integral becomes negative and how you have to split it into parts and take the absolute (positive) of certain quantities.
Post by: onur369 on July 21, 2011, 04:28:45 pm
Post by: onur369 on July 21, 2011, 10:20:51 pm
(http://i.imgur.com/wbZFT.jpg)
Post by: luken93 on July 21, 2011, 10:26:06 pm
hence integrate f(x) from 0->3 and let it equal 45 - this will allow you to determine a
From there, it's simply a matter of finding x ints to determine n and m.
Post by: tony3272 on July 21, 2011, 10:30:37 pm
VCAA 2006 EXAM1 Q11:
(http://i.imgur.com/wbZFT.jpg)
therefore the equation is
therefore
therefore m=6, n=-2 by the null factor law
Post by: onur369 on July 29, 2011, 11:04:22 pm
Post by: luken93 on July 29, 2011, 11:16:01 pm
Post by: tony3272 on July 29, 2011, 11:17:13 pm
Post by: xZero on July 30, 2011, 01:25:29 am
Post by: onur369 on August 03, 2011, 11:10:58 am
(http://i.imgur.com/N0uFf.jpg)
I can do a but not b :/
Post by: enwiabe on August 03, 2011, 01:42:27 pm
You get
You can now rearrange to arrive at
They now want you to find the anti-derivative of
Post by: onur369 on August 04, 2011, 07:30:40 pm
Any suggestions on how long I should take for Multiple Choice and Extended Response questions in Exam 2?
Post by: xZero on August 04, 2011, 09:08:34 pm
Post by: b^3 on August 04, 2011, 09:11:40 pm
Post by: vea on August 04, 2011, 09:21:49 pm
If they are actually CAS exams, then you're in a pretty good spot atm! :P
Post by: b^3 on August 04, 2011, 09:24:18 pm
Post by: onur369 on August 04, 2011, 09:26:41 pm
Post by: b^3 on August 04, 2011, 09:31:28 pm
Post by: pi on August 04, 2011, 09:32:33 pm
For ti nspire users: I have trouble finding exact values on my calculator even though the setting is set to exact. Suggestions?
On the calc page? Or the graph page?
I don't think it's possible on the graph page, but if the mode is on exact, the calc screen should always be displaying exact solutions (as long as there are no dodgy third or higher roots, or you use the decimal place in your input).
Post by: onur369 on August 04, 2011, 09:34:22 pm
Post by: pi on August 04, 2011, 09:36:30 pm
I get something like (12n1+pi)/12 :/
n1 = any integer
so sub in ... n= -1, n=0, n=1, n=3, ... etc. by hand, OR set a domain in your solve function (eg. -pi<x<pi) and it should give you all the solutions without the 'n' between that specified domain :)
Post by: b^3 on August 04, 2011, 09:36:48 pm
|0 equal to or less than x equal to or less than 2pi
with the proper signs in place of the words.
EDIT: beaten by 21 seconds.
Post by: xZero on August 04, 2011, 09:44:05 pm
Wait srsly only 10 mins on double checking? I've been finishing with like 1/2 hour to spare, is that too fast?thats just a general guideline, i always finish prac exam 1/2 an hour to spare as well, nearly had a heart attack during vcaa exam coz i thought i was too slow (only 10 min to double check)
Post by: b^3 on August 04, 2011, 09:48:20 pm
Post by: onur369 on August 10, 2011, 07:44:42 pm
Post by: onur369 on August 10, 2011, 07:45:22 pm
Post by: onur369 on August 16, 2011, 11:20:07 pm
No matter how many times I 'attempt or read' these types of questions I do not know what to do.
For example:
For the simultaneous linear equations
mx + 12y = 12
3x + my = m
find the value(s) of m for which the equations have
i. a unique solution
ii. infinitely many solutions.
ii. no solution
I do not know what to do for each, any help will be appreciated.
Post by: luken93 on August 16, 2011, 11:42:49 pm
From there;
unique solution = intersect once = different gradient, any c value.
infinitely many = always intersect = same gradient, same c value (same line)
no solution = never intersect = same gradient, different c (parallel)
Post by: onur369 on August 16, 2011, 11:59:19 pm
i.e: The simultaneous linear equations (m-2) x + 3y = 6 and 2x + (m - 3) y = m - 1 have no solution for
Post by: luken93 on August 17, 2011, 07:31:22 am
2x + (m - 3) y = m - 1 => y = -2x/(m - 3) + (m - 1)/(m - 3)
For it to have no solution, they must be parallel lines, ie the gradient must be the same:
-(m - 2)/3 = -2/(m - 3)
-(m - 2)(m - 3) = -6
-m^2 + 5m - 6 + 6 = 0
-m(m - 5) = 0
m = 0, m = 5
Now, we must also check to see that they are NOT the same line, so we must ensure that the 2 c values are not the same.
2 =/= (m - 1)/(m - 3)
2m - 6 =/= m - 1
m =/= 5
Therefore, we can't let it equal 5, because that would make it the same line. Hence sub in 0 to check the eq:
=> y = 2x/3 + 2
=> y = 2x/3 + 3
The two lines are parallel, but not the same line, hence they will never intersect.
Post by: onur369 on August 23, 2011, 11:51:17 pm
Q: A kindergarten teacher has found over the years that 25% of children can tie their shoelaces, and 30% can use a pair of scissors, and 18% can do both. Find the probability that a randomly selected child:
a) can neither ties their shoelaces nor use scissors
b) can use scissors, but cannot tie their shoelaces
for a) i tried doing 0.7x0.75= 0.525, but the solution is 0.63.
Post by: luken93 on August 23, 2011, 11:57:24 pm
Pr(Scissors) = 0.30
Pr(Shoe n Scissors) = 0.18
Since Pr(Scissors u Shoes) = 0.30 + 0.25 - 0.18 = 0.37
Neither = 1 - Pr(Scissors u Shoes) = 1 - 0.37 = 0.63
b) Pr(Scissors only) = Pr(Scissors) - Pr(Scissors n Shoes) = 0.30 - 0.18 = 0.12
Post by: kaushik on August 24, 2011, 06:56:31 am
Post by: onur369 on September 30, 2011, 03:31:05 pm
Show that g(u + v) + g(u – v) = 2(g(u) + g(v)).
How do i do this ?
Post by: tony3272 on September 30, 2011, 03:48:00 pm
Let LHS=
Post by: onur369 on October 01, 2011, 11:35:50 am
this question is really weird for me :s:
Show that the graph of h(x)= x^n/e^x , where n is a positive integer, has a local maximum at x=n.
Post by: b^3 on October 01, 2011, 11:44:04 am
Take an e^x out
Now let this equal zero to find maximum and minimums
so
Then show it is a maximum by subbing in terms to the left and right, i.e. n-1 and n+1, as n is postive, you can should that the left side will have a positive gradient and right side has a negative gradient, hence a maximum.
The gradient on the left at n-1 is
As n is a postive interger, n-1 is also positive and n/n-1 is greater than 1 so the gradient on this side is positive
Now on the right at n+1 the gradient is
As n+1 is greater than n, n/n+1 is a fraction smaller than one and greater than zero, so if you take the one away, you get a negative. The other terms are still positive so overalll it multiplies to a neagtive.
There is proabably a more mathematical way to write this out, so if anyone wants to add to it (and well do it the proper way) feel free to.
EDIT: added last bit of working.
Post by: onur369 on October 01, 2011, 11:52:27 am
Post by: b^3 on October 01, 2011, 11:54:52 am
Post by: onur369 on October 11, 2011, 07:18:10 pm
Post by: b^3 on October 11, 2011, 07:36:45 pm
a)
Now you need to use the ratios of the similar triangles.
Sub that in
b)
That seems odd, let me have a check of it.
Post by: onur369 on October 11, 2011, 07:43:11 pm
Post by: onur369 on October 16, 2011, 09:51:02 pm
a: 0.6745
b: -0.6745
c: 0.5987
d: 0.7734
e: -0.7734
Post by: tony3272 on October 16, 2011, 09:56:56 pm
Just using inverse normal on your CAS:
-c = invnorm(0.75)
-c=0.6745
therefore c=-0.6745
Post by: onur369 on October 16, 2011, 10:00:41 pm
Post by: onur369 on October 18, 2011, 04:31:43 pm