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| Author | miscellaneous |
|---|---|
| Tags | author:miscellaneous question unrated |
| Created | 2007-11-18 |
| Last Modified | 2007-11-19 |
| Rating | 4 more votes required for a rating. |
| Map Data | |
| Description | ohhh....
the ninja was run over use/credit "stop the abuse," cry the tilesets THIS IS NOW PART OF THE QUESTION OF THE DAY SERIES!!! i'm gonna be a complete düsh (fun to spell like that) and make it a math problem :( what is the slope of line: |2x-7|=2y+7 SCOREBOARD gforce20: 1 pepi321: 1 everyone else: 0 |
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| SIMPLE CHALLENGE | Bird of Prey | response to Fire Tamer | QUESTION OF THE DAY | QUESTION OF THE DAY 2 | hitman logo |
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2007-11-25
Look miscellaneous...
Look carefully at the question yourself. Just read it. Then, you will realise it is an absolute value graph. HOW DOES THE ABSOLUTE VALUE GRAPH LOOK?
After that, reconsider your answers.
I think more and more people are waking up to this.
After that, reconsider your answers.
I think more and more people are waking up to this.
2007-11-24
Misc:
I suggest you check your working. What did your graph come out as?
If I am wrong, I am interested to see how.
If I am wrong, I am interested to see how.
2007-11-24
McFizzle:
Finding two points isn't exactly proof. sub x=5 and you'll find that y=-2.
I did the working and I get:
when {x<3.5; x=-y} therefore grad is -1
when {x>3.5; x=y+7} therefore grad is 1
and at x=0, there is no soln to grad because it is a critical point.
I did the working and I get:
when {x<3.5; x=-y} therefore grad is -1
when {x>3.5; x=y+7} therefore grad is 1
and at x=0, there is no soln to grad because it is a critical point.
2007-11-20
I can prove it's actually -1
If you substitute in 1 for x, you will get y = -1.
If you substitute in 2 for x, you will get y = -2.
This leaves you with the points (1, -1) and (2, -2). Graph those points, and the slope is in fact -1. This works with any number you substitute in for x. The slope is -1 and the line passes through the origin.
If you substitute in 2 for x, you will get y = -2.
This leaves you with the points (1, -1) and (2, -2). Graph those points, and the slope is in fact -1. This works with any number you substitute in for x. The slope is -1 and the line passes through the origin.
2007-11-20
well,
i asked both my sisters calculus teacher and my college professor mother, and they said 1
2007-11-20
It's seriously both.
I'm in australia, i go to a selective school (smart) and they all said both when i posed the question.
SO.
SO.
2007-11-20
actually...
i think i may have messed up, because i dont think you can divide 2 into an absolute value... so.
|2x-7| = 2
2x-7 = 2 and 2x-7 = -2
2x = 9 and 2x = 5
x = 9/2 and 5/2.
oh ok... i get the same answers just i put a negative where i shouldnt have... oops. so this is my answer.
slopes = 9/2 and 5/2.
|2x-7| = 2
2x-7 = 2 and 2x-7 = -2
2x = 9 and 2x = 5
x = 9/2 and 5/2.
oh ok... i get the same answers just i put a negative where i shouldnt have... oops. so this is my answer.
slopes = 9/2 and 5/2.
2007-11-20
lets see...
its a little late and it wont matter but who cares...
ok first of all, always work with the problem first, because there may not always be 2 solutions.
next you subtract 7 from the side with y on it to get,
|2x-7|-7=2y.
from there you divide by 2. which would give you
|x-(7/2)|-(7/2)= y
and now you want to isolate x on 1 side now that y=1.
solve the absolute value,
|x-(7/2)|= 1, because y = 1.
x-(7/2)=1 and x-(7/2)=-1
x = -9/2 and x = 5/2
the slopes are both -9/2 and 5/2
am i right? i doubt it. i just tried to explain everything the way me teacher taught me and im in my schools matheletes so idk. if im wrong tell me how and where. congrats to those who won this.
ok first of all, always work with the problem first, because there may not always be 2 solutions.
next you subtract 7 from the side with y on it to get,
|2x-7|-7=2y.
from there you divide by 2. which would give you
|x-(7/2)|-(7/2)= y
and now you want to isolate x on 1 side now that y=1.
solve the absolute value,
|x-(7/2)|= 1, because y = 1.
x-(7/2)=1 and x-(7/2)=-1
x = -9/2 and x = 5/2
the slopes are both -9/2 and 5/2
am i right? i doubt it. i just tried to explain everything the way me teacher taught me and im in my schools matheletes so idk. if im wrong tell me how and where. congrats to those who won this.
2007-11-19
@ dAaZnfr3aK
umm... no it was to the first, emphasis on first, one to get it. also, the answer cannot be what you said because:
the appropriate answer, solving the absolute value first, would be y=x
so, intercepts are both (0,0) which means m=y2-y1/x2-x1 m=0-0/0-0 or m=0/0 anything divided by itself, even 0, is 1
so, in conclusion, as Hendor was first, he gets 1/10 of a ded, you were wrong, so you get nothing. joshwapiano may get a consolation prize, maybe on my next map.
the appropriate answer, solving the absolute value first, would be y=x
so, intercepts are both (0,0) which means m=y2-y1/x2-x1 m=0-0/0-0 or m=0/0 anything divided by itself, even 0, is 1
so, in conclusion, as Hendor was first, he gets 1/10 of a ded, you were wrong, so you get nothing. joshwapiano may get a consolation prize, maybe on my next map.
2007-11-19
NO IT DOES NOT GO TO HENDOR.
JOSH WA PIANO IS CORRECT. OR ME.
2007-11-19
if you solve the equation...
you get 2 answers.
y=x
or y=x-7.
discarding the 1st one for an obvious reason...
sub x = 0,
y intercept is 0,7
x intercept is 7,0
gradient (m) = y2-y1/x2-x1
therefore= 0-7/7-0
=-1.
but
its absolute value, you could accept either one.
correct me if i'm wrong.
y=x
or y=x-7.
discarding the 1st one for an obvious reason...
sub x = 0,
y intercept is 0,7
x intercept is 7,0
gradient (m) = y2-y1/x2-x1
therefore= 0-7/7-0
=-1.
but
its absolute value, you could accept either one.
correct me if i'm wrong.
2007-11-19
this is an absolute value graph.
it can't have a gradient. however, thereis a gradient for 1 side
2007-11-19
hang on...
do you even know what the slope is?
2007-11-19
wow...
goes to hendor
omfg, no one in my school knew that
omfg, no one in my school knew that
2007-11-19
The answer is
+1 or -1, both exist in graph of the function.
Because of the modulus signs, the graph should look like a V.
Because of the modulus signs, the graph should look like a V.
2007-11-18
ok
the slope is 1 isnt it?
cause the absolute value thing means that you can change that negative to a positive, so if you have
2x + 7 = 2y + 7
then x and y have to be the same thing. which could be one or 18 or 72. so ya. my answer is: the slope is one.
and dont call me a nerd or something, cause i got lucky; we just did this in my algebra class.
cause the absolute value thing means that you can change that negative to a positive, so if you have
2x + 7 = 2y + 7
then x and y have to be the same thing. which could be one or 18 or 72. so ya. my answer is: the slope is one.
and dont call me a nerd or something, cause i got lucky; we just did this in my algebra class.
2007-11-18
answer's 1
what's the point of the map though... knida useless.
| Demo Data |
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2007-11-18
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peace_bear0
Is this right?
Here's my work:
|2x-7|=2y+7
2x-7 = 2y+7 AND -2x+7 = 2y+7
/2,-7 /2,-7 /2,-7 /2-7
x-14=y AND x=y
y=x-14 AND y=x