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#51 Re: Help Me ! » Limits » 2006-10-26 12:15:16
oh woops, my window setting were too small, i didnt see the other part of the function.
#52 Re: Help Me ! » Limits » 2006-10-25 18:02:07
2)
Here the limit is not finite. As x approaches -3 from -'ve infinity the given function approaches -infinity
so:
when I graph it, it looks like as it approaches -3 from the negative side, it rises to positive infinite, not negative.
#53 Re: Help Me ! » Simple Calculus » 2006-10-25 17:57:54
so can you get the points that are discontinuous by simply finding what x cannot equal? For f(x) = (x-2)/(x² -4), and got 2(removable because it's smooth) and -2(non-removable because it goes off into infinite). Is this right or am I misunderstanding you guys?
#54 Help Me ! » Simple Calculus » 2006-10-25 12:07:46
- fusilli_jerry89
- Replies: 4
Fine the values of x for which f(x) = (x-2)/(x²-4) is discontinuous and label these discontinuities as removable or unremovable.
I just simple do not get what these words means, removable, unremovable and discontinuous. I can guess but I wanna be sure.
#55 Re: Help Me ! » Another Physics Problem » 2006-10-25 12:05:42
so basically they are the same but one is negative?
#56 Re: Help Me ! » Another Physics Problem » 2006-10-24 18:08:26
what is the difference between the normal force and what you call Fp?
#57 Help Me ! » Fireman Problem » 2006-10-23 07:50:30
- fusilli_jerry89
- Replies: 4
A fireman has a hose and is standing on top of a 20 m high building. His hose shoots water at 12 m/s and he wishes to hit the top of another 20 m hig building 21 m away. what angle should he aim his hose at?
I calculated that in the x direction, the velocities are both 12cosθ, the acceleration is zero, distance is 21.0 and te time is unknown. In the y direction, te initial velocity is 12sinθ and the final is -12sinθ, the acceleration is -9.8, the distance is zero, and the time is also unknown.
I then set up a system:
x: t=21/12cosθ
y: t=24sinθ/9.8
0=21/12cosθ-24sinθ/9.8
(21/12)secθ=(24/9.8)sinθ
(21/12)/(24/9.8)secθ=sinθ
sinθ/secθ=205.8/288
sinθcosθ=205.8/288
I'm stuck now...
#58 Help Me ! » Limits » 2006-10-22 18:38:06
- fusilli_jerry89
- Replies: 9
lim (2- √(6-x))/(x-2)
x->2
lim (3x²-x+1)/(x+3)
x->-3-
lim (√(x+Δx)- √(x))/Δx
Δx->0
lim (sin3x)/x
x->0
lim (x+2)/(x³+8)
x->-2
Without graphing
#59 Help Me ! » Physics again(sorry) » 2006-10-18 07:34:57
- fusilli_jerry89
- Replies: 1
Sorry no toher forum is wroking rightnow so i am forced to post here.
a 2.0 kg brick rides on a 20kg sled and is dragged upa22 degree slope by a rope witha tension of 1000 N. Theropeis parallel totheslope. The friction coefficient(both static and kinetic) between the brick and sled is 0.30 and between the sled and the ground is 0.15. Predict each mass's acceleration. I got 0.8 m/s/s for thebrick which i'm sure is right, and for the sled sled I got 40 m/s but the teacher says it is 45 m/s/s.
Teacher's method:
1000-20(9.8)sin22-0.15(22)(9.8)sin22-5.4=20a
a=45.5m/s/s
My method:
1000-81-30-5.4=22a
a=40m/s/s
#60 Re: Help Me ! » Another Physics Problem » 2006-10-17 17:03:31
wat did u do between your last two posts? bcuz u got the right answer
#61 Re: Help Me ! » Easy Physics » 2006-10-17 16:41:06
is there a way to get the Force of friction without using conservation of energy? like just newton's laws for example?
#62 Re: Help Me ! » equation of the line » 2006-10-14 09:27:20
#63 Re: Help Me ! » simplify......am i right? » 2006-10-12 12:49:16
unique wrote:3 - sqrt(2)
------------
4 + 2sqrt(2)
for the very beginning part when u times the fraction by the "negative" version of the denominator, is that technique only used for this stuff? It seems if you use that on any other question itll work also.
#64 Help Me ! » Questions I forgot how to do » 2006-10-12 12:22:04
- fusilli_jerry89
- Replies: 1
1) Factor:
x^4-81y^8
wait this one must be (x^2+9y^4)(x^2-9y^4)
2) You have two large containers. The first contains 9% solution and the second 12%. To make 4 litres of 10% solution, how many litres of each solution should you use?
I did this and ended up getting 12.67 and -8.67. How can it be negative?
#65 Help Me ! » Another Physics Problem » 2006-10-04 10:59:23
- fusilli_jerry89
- Replies: 15
Sorry for posting another physics problem here, but you are the only ones that actually help.
The drawing shows a large cube(mass=25 kg) being accelerated accross a frictionless surface by a horizontal force P. A small cube(mass=4.0 kg) is in contact with the front surface of the large cube and will slide down unless P is sufficiently large. The coefficient of static friction between the cubes in 0.71. WHat is the smallest magnitude that P can have in order to keep the small cube from sliding downward?
I understand that an extra 11.368 Newtons must be directed upwards to keep the block in place but I don't know what to do after that.
#66 Help Me ! » Easy Physics » 2006-10-02 11:17:47
- fusilli_jerry89
- Replies: 3
I don't know if anyone solves physics problems here, but I havn't done this in over a year and can't concentrate right now so I need help.
A 1.2 kg book is given a shove at 2.0 m/s. It slides across the table coming to a stop ).67 metres later. a) what is the magnitude of the force that stops the book? b) What is the coefficient of kinetic friction between the book and table and c) If released at the same speed, would a heavier book slide the same distance? Explain
#67 Re: Help Me ! » Graphing » 2006-09-20 16:37:55
I already know how to get those answers, but the answers in the back of the book are as follows:
a) x = -1 + 5t, y = -3 + 4t, 0≤t≤1
b) x = t² + 1, y = t, t≤0
#68 Help Me ! » Graphing » 2006-09-20 14:55:31
- fusilli_jerry89
- Replies: 2
Fine a parametrization for the curve:
a) the line segment with endpoints (-1,-3) and (4,1)
b) the lower half of the parabola x - 1 = y²
#69 Help Me ! » More Logarithms » 2006-08-13 15:46:41
- fusilli_jerry89
- Replies: 1
is there a way to simplify a log equation with different bases?
i.e. log7base8 - log5base4
I can only get it down to: (log7/3log2)-(log5/2log2)
#70 Re: Help Me ! » couple ?'s » 2006-08-13 15:11:34
I found another way to do 45 is what % of 75.
45 + 30 = 75
15 + 10 = 25 because I divided everything by 3.
Add the last two equations together.
45 30 75
+15 +10 +25
----- ----- -----
60 40 100Yeah 60% + 40% = 100%
We did it!!!!!!!!!!!!!
uhh, or you could just divide 45 by 75 which gives you 0.6.....
and for the second question, just write a simple equation, 0.75x=36, which means 75% of a number(x) equals 36. Solve and you get 48.
#71 Help Me ! » Logarithms » 2006-08-13 11:45:51
- fusilli_jerry89
- Replies: 1
If log8=m and log7=n, write each logarithm in terms of m and n:
log(8/49) [fraction, not base]
for this, i put log8-log7-log7 which is m-2n, but the teacher marked it wrong.
log448
for this one, i got 2log8+log7, which is 2m+n, but once again, the teacher marked it wrong.
Are these answers right, or is the teacher wrong?
#72 Help Me ! » combinatorics » 2006-08-12 11:55:25
- fusilli_jerry89
- Replies: 2
Determine the total number of possible "words" using at least one of the letters of B-A-R-H-A-M.
I know how to get all the 6 letter words, 6!/2!, and the 1 letter words, obviously 5, but is there an easier way to get all the 2,3,4 and 5 letter words instead of just counting?
#73 Re: Help Me ! » How to solve this eqation? » 2006-08-12 09:12:39
where does the 'e' dissapear off to?
#74 Re: Help Me ! » Simple Trig » 2006-08-12 08:59:51
k well thanks then
#75 Re: Help Me ! » need help on a simple log question » 2006-08-11 23:32:47
53=10 log (1/T)
log 1/T = 5.3
10^5.3=1/T
199526.2315=1/T
T=5.011872336 x10^-6
25=10 log (1/T)
log 1/T = 2.5
10^2.5=1/T
316.227766=1/T
T=0.0031622777
20=10 log (1/T)
log 1/T = 2
100=1/T
T=0.01
or of the numerators meant bases:
53=10 log (1/T)
log 1/T = 5.3
log T/Log 1=5.3
log 1 = 0;denominators can't be zero(same for all)
or if the denominators meant bases:
53=10 log (1/T)
log 1/T = 5.3
log 1/log T = 5.3
log 1=0
0/log T=5.3
there is no answer in this case