^nope, definitely not punching in wrong numbers ^_^
http://s4.postimage.org/phkf063zw/Snapshot_20111010.jpg
perhaps it's a calculator error? i really want to know because there's a test coming up...
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hmm~
Member
10:21 pm, Oct 9 2011
Posts: 989
Mishy
Member
10:30 pm, Oct 9 2011
Posts: 1737
Quote from MewMan
^nope, definitely not punching in wrong numbers ^_^
http://s4.postimage.org/phkf063zw/Snapshot_20111010.jpg
perhaps it's a calculator error? i really want to know because there's a test coming up...
http://s4.postimage.org/phkf063zw/Snapshot_20111010.jpg
perhaps it's a calculator error? i really want to know because there's a test coming up...
Try putting a division sign between the 2 and 3. To my knowledge, the sign you used is only for improper fractions, but I could be wrong. I use a TI-84. But even if you don't use a calculator, you can see that the cube root of (-1) is (-1) and (-1)^2 is (1), right?
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hmm~
Member
10:34 pm, Oct 9 2011
Posts: 989
yeah, i know it's 1
it just bugs me that there's a chance of sudden death on a test because of calculation errors
and btw, division sign does not work
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it just bugs me that there's a chance of sudden death on a test because of calculation errors
and btw, division sign does not work
________________
its cold down here fam ~
Mishy
Member
10:41 pm, Oct 9 2011
Posts: 1737
Quote from MewMan
yeah, i know it's 1
it just bugs me that there's a chance of sudden death on a test because of calculation errors
and btw, division sign does not work
it just bugs me that there's a chance of sudden death on a test because of calculation errors
and btw, division sign does not work
I'm sorry I can't help you further. I would recommend that you get a TI-83 or TI-84 in the future for your Calculus courses. Anything above your teacher may not allow.
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Nulla in mundo pax sincera
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Member10:41 pm, Oct 9 2011
Posts: 2050
/Creeps behind wall..... wants to see pictures of MewMan.... |ω･）
What class is this for, sir MewMan?
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What class is this for, sir MewMan?
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Quote from Crenshinibon
And you will murder someone one day, pika. If you're my daughter.
hmm~
Member
10:43 pm, Oct 9 2011
Posts: 989
thank you for your help 
it's too bad but we are not supposed to use a graphic calculator on the test T_T
@Pikapu: Calculus I
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it's too bad but we are not supposed to use a graphic calculator on the test T_T
@Pikapu: Calculus I
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its cold down here fam ~
Mishy
Member
10:45 pm, Oct 9 2011
Posts: 1737
Quote from MewMan
thank you for your help
it's too bad but we are not supposed to use a graphic calculator on the test T_T
@Pikapu: Calculus I
it's too bad but we are not supposed to use a graphic calculator on the test T_T
@Pikapu: Calculus I
Just remembered, I'm not allowed either!
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Nulla in mundo pax sincera
"Always go too far, because that's where you'll find the truth." - Albert Camus
non-standard
Member
11:01 pm, Oct 9 2011
Posts: 740
I have the older version of that calulator.
The only way around it is to split the fraction.
By that I mean type
( (-1)^2 ) ^(1/3) ) or
( (-1 )^(1/3) ) ^2
I threw in some extra spaces because I was getting confused. Hope it makes sense.
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The only way around it is to split the fraction.
By that I mean type
( (-1)^2 ) ^(1/3) ) or
( (-1 )^(1/3) ) ^2
I threw in some extra spaces because I was getting confused. Hope it makes sense.
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Site Admin
3:49 am, Oct 10 2011
Posts: 2275
You can't use the Mean value theorem because x^(2/3) over the interval (-1,1) is not a smooth curve. It looks like a v.
For mean value theorem, the curve has to be smooth.
(are you not allowed to use the power rule... or is that cheating at your stage?)
It would show you that from -1=>0 d/dx of x^(2/3) is -2/3 and from 0=>1 d/dx of x^(2/3) is 2/3 therefore over the interval DNE... I think... or it's just not differentiable over the interval. Of course, if you are allowed to split it up, you'd be fine... but I don't think that's the case for you.
might want to get to know http://www.wolframalpha.com
Last edited by Toto at 4:22 am, Oct 10 2011
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For mean value theorem, the curve has to be smooth.
(are you not allowed to use the power rule... or is that cheating at your stage?)
It would show you that from -1=>0 d/dx of x^(2/3) is -2/3 and from 0=>1 d/dx of x^(2/3) is 2/3 therefore over the interval DNE... I think... or it's just not differentiable over the interval. Of course, if you are allowed to split it up, you'd be fine... but I don't think that's the case for you.
might want to get to know http://www.wolframalpha.com
Last edited by Toto at 4:22 am, Oct 10 2011
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and I don't know you. Make sure it doesn't leave the building."
Member
10:47 am, Oct 10 2011
Posts: 38
I'm not a native english speaker so I hope someone can help me:
my friend asked me for advice if she should write "made by artist (name)" or "made by the artist (name)"
I want to answer her question but I'm not sure...
which phrase is right? (:
my friend asked me for advice if she should write "made by artist (name)" or "made by the artist (name)"
I want to answer her question but I'm not sure...
which phrase is right? (:
Member10:54 am, Oct 10 2011
Posts: 2050
Either way would work.
You could do, "made by [artist name]." or "made by the artist, [artist name]." In the second version, the comma is necessary.
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You could do, "made by [artist name]." or "made by the artist, [artist name]." In the second version, the comma is necessary.
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Quote from LawX
You are like the dense main character in a shoujo manga.
Quote from Crenshinibon
And you will murder someone one day, pika. If you're my daughter.
Member
11:03 am, Oct 10 2011
Posts: 38
Quote from Pikapu
Either way would work.
You could do, "made by [artist name]." or "made by the artist, [artist name]." In the second version, the comma is necessary.
You could do, "made by [artist name]." or "made by the artist, [artist name]." In the second version, the comma is necessary.
thank you very much
Mishy
Member
7:49 pm, Oct 10 2011
Posts: 1737
I've got a math question:
A perfect cube is an integer whose cube root is an integer. For example, 27, 64 and 125 are perfect cubes. If p and q are perfect cubes, which of the following will not necessarily be a perfect cube?
A. 8p
B. pq
C. pq + 27
D. -p
E. (p - q)6
I know the right answer is C, but I want to know why E cannot be the answer. The answer given by the question site is
Last edited by chineserider at 8:02 pm, Oct 10 2011
________________
Nulla in mundo pax sincera
"Always go too far, because that's where you'll find the truth." - Albert Camus
A perfect cube is an integer whose cube root is an integer. For example, 27, 64 and 125 are perfect cubes. If p and q are perfect cubes, which of the following will not necessarily be a perfect cube?
A. 8p
B. pq
C. pq + 27
D. -p
E. (p - q)6
I know the right answer is C, but I want to know why E cannot be the answer. The answer given by the question site is
Quote
Since the difference between p and q is raised to the power of 6, this expression will be a cube (with cube root = difference squared).
But I have no idea what that means. Would someone please explain it to me?Last edited by chineserider at 8:02 pm, Oct 10 2011
________________
Nulla in mundo pax sincera
"Always go too far, because that's where you'll find the truth." - Albert Camus
hmm~
Member
8:17 pm, Oct 10 2011
Posts: 989
Quote from chineserider
I've got a math question:
E. (p - q)6
I know the right answer is C, but I want to know why E cannot be the answer. The answer given by the question site is
E. (p - q)6
I know the right answer is C, but I want to know why E cannot be the answer. The answer given by the question site is
Quote
Since the difference between p and q is raised to the power of 6, this expression will be a cube (with cube root = difference squared).
But I have no idea what that means. Would someone please explain it to me?shouldn't E be (p-q)^6 then?
just to be clear: (p-q)^6=((p-q)^2)^3= cube
p, q =perfect cube=integer =>p-q= integer
=>(p-q)^6= perfect cube
Last edited by MewMan at 8:23 pm, Oct 10 2011
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Mishy
Member
7:26 am, Oct 11 2011
Posts: 1737
Quote from MewMan
Quote from chineserider
I've got a math question:
E. (p - q)6
I know the right answer is C, but I want to know why E cannot be the answer. The answer given by the question site is
E. (p - q)6
I know the right answer is C, but I want to know why E cannot be the answer. The answer given by the question site is
Quote
Since the difference between p and q is raised to the power of 6, this expression will be a cube (with cube root = difference squared).
But I have no idea what that means. Would someone please explain it to me?shouldn't E be (p-q)^6 then?
just to be clear: (p-q)^6=((p-q)^2)^3= cube
p, q =perfect cube=integer =>p-q= integer
=>(p-q)^6= perfect cube
Ah, I see. I was thinking about it the wrong way. Thanks
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Nulla in mundo pax sincera
"Always go too far, because that's where you'll find the truth." - Albert Camus
