Printable View
There is a question that I can never figure out why it is wrong. Here is the question:
16-36=25-45
= -20
16-36+20.25=25-45+20.25
add 20.25 to both sides
FOIL / Complete the square...
(4-(9/2))^2 = (5-(9/2))^2
Sqrt it.
4-(9/2) = 5-(9/2)
add both sides by 9/2
4=5
why is it wrong?
This question held me prisoner for 5 hours and I just lost my last bit of hand-eye coordination.
---------------------------
Engineer223
well, I'll take a look at it, but I'm sure that the problem is with the sqrt.
you see, you can't do
a^2=b
a=sqrt(b)
that's not true, the solution is
a^2=b
a=(+/-)sqrt(b)
that's because 5^2=(-5)^2=25
but I'll give you the full answer in a while
Oskilian
the trick is that when you square it, it can remove the negative... when you square it you are supposed to take the negative on the four and the positive on the five, thus giving you the difference... [+-0.5]
yup, that's it
Oskilian
hasafraggin 3 minutes...
What are you studying? I mean what grade.
kindergarten.
j/k:)
LOL, but don't be rude ;)
Haha, I got the a similar problem once...
y=root X (can't type the root sign)
is a function
y^2=X
(square both sides)
is not a function...
y^2=X is a function.
What is the answer to (10˛ + 11˛ + 12˛ + 13˛ + 14˛) / 365 WITHOUT using your calculator.
Hint: {}
--------------------------
Engineer223
Midterm on Friday!!! ahhhh!!! [banging my head on the wall]
simple...2 :-)
and no I didnt use a calculator.
no it isn´t, a function must have only one X value for each Y value, and this one doesn´t (for example: if X=4, y can be 2 or -2)Quote:
Originally posted by Gades
y^2=X is a function.
Oskilian
Ok here is one for you all.
y^x = x^y
x and y are not equal.
x=4
y=2
--
yeah.
I was trying with logarithms :rolleyes:
Oskilian
2 and 4 are correct. Now whats the second set of numbers?
>>no it isn´t, a function must have only one X value for each Y >>value, and this one doesn´t (for example: if X=4, y can be 2 >>or -2)
Osk, you're trying to writ some code where you need to understand 3D. OK, that's fair. But, then admit that you can have more than one value of each X and Y.
For example, let's say I want to define a circle in 3D (even the circle has only 2D, it's in a space with 3D):
1) x^2 + y^2 = 1
2) z = 0
The ecuation 1) defines the limits (I don't know how to express this in Enlgish) of the function in the X-Y plane, and the function 2) express the position of the circle in the Z axis.
And the function Y^2 = x is used many times when you need a 3D boundary to calculate for example triple integrals, which give you the volume of a body.
By the way, I took this examples from my first year at uni (5 or 6 years ago I think).
And these functions are quite simple, because a value of x has only two values of y; there are functions that have more y values for each x value. Like the Rosae Function, this function draws like a rose, but I don't know if you call it like this in English as well. If I find something in my notes I'll post it.
x = 4Quote:
Originally posted by Thantos
2 and 4 are correct. Now whats the second set of numbers?
y = 2
really, thats trivial
/\
|
That the same one as mine first answer
x=2
y=4
--------------------------
engineer223
No re-read my comment. I said 2 and 4 were correct. But there is another correct set of numbers. Meaning that 2 and 4 isn't the only set of correct numbers.
Is it -2 and -4
Yeppers
a circle is not a function, it's an equation, but not a function
Oskilian
y = sqrt(x)
It passed the vertical line test so it is a function
Next time, make "y" alone before calculating.
-------------------
Engineer223
GEEZ! Engine, you don't get my point...
> y˛ = x is a function
sqrt both sides
y = sqrt(x)
It passed the vertical line test so it is a function<
That was exactly what I said, instead I said it the other way around... You just proved my point, y sqrt(x) is a function while y^2=x is exact same equation but not a function. ONLY y = sqrt(x) is a function the other one is not. The definition for a function, is that it's linear graph thing is not touched twice by a vertical line. And we are talking about 1D here ( is there such a thing as 1D?...) ...
the thing with your examples is whether to decide to take the negative portion of a sqrt... and it's just a matter of orientation...
you must undertand that you cannot do that!
the inverse function of the second power IS NOT int sqare root!!!
graphing y^2=x will give you a different graph than y=sqrt(x)
now this is true:
y^2=x
y=(+/-)sqrt(x)
but neither of them are functions, let me copy the "function" definition from my Calculus book
one UNIQUE element; in the equation y^2=x, y=2 and y=-2 are assigned to x=4. There are two solutions for the eqation, and THAT makes y^2 a non-functionQuote:
Originally written by Earl W. Swokowsky
An f function of a D group over an E group is a correspondence which assigns each element x of D one unique element y of E
YES!!! urrrggghh....
you see if you start with
y=ROOT(x)
that IS a function.
but if you move the root sign over, you will get y^2=X
which IS NOT a function. (because x will be both -2 and 2)
YOU DON'T START WITH y^2=x, ok?
Do I have to graph it?
Proof
y=sqrt(x)
---------------------
Engineer223
YES!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
OMG!!!
YOU DON'T READ DO YOU???
urrrr....
yes, y=sqr(x) IS A FUNCTION. BUT I NEVER TALKED ABOUT y=sqr(x). I TALKED ABOUT y=ROOT(x)... geez... and I only said y^2=x is NOT a function. Get it? I hope so... *sigh* is my english bad or what...
ironically enough, x ^ 2 + y ^ 4 = 2 ^ 2 + 2 ^ 4, find two sets of values for x and y, was coincidentally a problem i had today in a math contest... thanks! actually i figured it out then realized it was stated here, but anyways... it was fun to be in those again...
sqrt() = ROOT
sqrt() = sq rt!!!
Square ROOT!!!!
..._
0.9 ? 1.00
Not equal or equal?
Being posted in the SC S&T forum b4.
-------------------------
Engineer223
P.S. Michelle called me an engine!:mad:
0.9 is not equal, never has been equal and never will be equal to 1.