mathmath333
  • mathmath333
How to graph this.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} \mid y\mid \geq 1\hspace{.33em}\\~\\ \end{align}}\)
Vocaloid
  • Vocaloid
\[\left| y \right| \ge 1 \] means that \[y \ge 1\] or \[y \le -1\] |dw:1434816235241:dw|
mathmath333
  • mathmath333
u mean this one |dw:1434816613473:dw|

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More answers

Vocaloid
  • Vocaloid
yeah, that graph is lot neater, thanks!
mathmath333
  • mathmath333
@Vocaloid but wolfram gave this as the graph |dw:1434816760211:dw|
hartnn
  • hartnn
1. |y| >= 1 is a one variable inequality. so you'd use a number line. 2. If you want a xy plot, your plot should show regions for all values of y... like if y = -2, would your inequality satisfy ?? is it shown by that plot?
hartnn
  • hartnn
point 1,2 satisfies your inequality, but the wolf graph says it doesn't :)
hartnn
  • hartnn
wolf graph is wrong. it plots x =|y| and y =1
mathmath333
  • mathmath333
how can i plot 1 in (1,2) there is no space for x co-ordinate
hartnn
  • hartnn
it should have plotted x=1 instead of y=1
hartnn
  • hartnn
|dw:1434817459914:dw|
mathmath333
  • mathmath333
|dw:1434817543444:dw|
ganeshie8
  • ganeshie8
how do you sketch y = 1 in 1D, 2D and 3D ?
mathmath333
  • mathmath333
|dw:1434817611958:dw|
hartnn
  • hartnn
the wolf graph took vertical axis as 'x' ! lol
hartnn
  • hartnn
if you want a 2D graph for |y|>=1, then the first graph is correct
ganeshie8
  • ganeshie8
Haha wolfram is trolling us |dw:1434817710812:dw|
hartnn
  • hartnn
|y|≥1 means that y≥1 or y≤−1 is absolutely correct
mathmath333
  • mathmath333
but why is that desmos graphing calculator also cannot graph \(|y|\geq 1\) when i input the syntax.
ganeshie8
  • ganeshie8
As the question stands, nobody can stop me from thinking of \(y\) as distance from origin (polar) ;p |dw:1434817996160:dw|
hartnn
  • hartnn
"We don't solve complicated single variable equations yet" lol true ganeshie! it depends on how we want to graph...multiple correct answers.
hartnn
  • hartnn
although the best thing to do here is to plot on the number line!
mathmath333
  • mathmath333
so this below graph correct for \(|y| \geq 1\) |dw:1434818180900:dw|
ganeshie8
  • ganeshie8
yeah some are ridiculous and some are more or less obvious from the context
hartnn
  • hartnn
yes, it is math. verify by taking different values of y
mathmath333
  • mathmath333
ok thnx
Vocaloid
  • Vocaloid
hm, I suppose graphing calculators/graphing software is usually designed to handle x and y together think of it this way: we can graph y = 1, right? it's just a horizontal line through y = 1 y > 1 is just everything shaded above that line.
hartnn
  • hartnn
y=1 is a point on a number line. y= 1 is a line in xy plot: 2D y=1 is a plane in xyz plot: 3D
Vocaloid
  • Vocaloid
^ that, too :)
mathmath333
  • mathmath333
i have checked in geogebra too, it also can't plot \(|y|\geq 1\)
ParthKohli
  • ParthKohli
But you can.
ganeshie8
  • ganeshie8
try below in geogebra abs(y) >= 1
hartnn
  • hartnn
so you > desmos/geogebra :P :)
ParthKohli
  • ParthKohli
We created Desmos and Geogebra. Humans 1 - 0 Computers
mathmath333
  • mathmath333
tried this "abs(y) >= 1" , nothing happened from geogebra
mathmath333
  • mathmath333
after all man made computer , computer did'nt made man
ganeshie8
  • ganeshie8
id love to be in a matrix for a change
hartnn
  • hartnn
|y| >= 1 the first thought that came to mind, all values of y, that are greater than 1, irrespective of the sign of the number....
ParthKohli
  • ParthKohli
There's really not much to think about this problem. You just find the set of points with the absolute value of their y-coordinate either 1 or greater than that. So the "base"-set of points would be points with y-coordinate = 1. I just realised that it's exactly what hartnn explained but I don't want to remove this reply so yeah
hartnn
  • hartnn
|dw:1434819148818:dw|
ganeshie8
  • ganeshie8
|dw:1434819259110:dw|
ParthKohli
  • ParthKohli
|dw:1434819280155:dw|
ganeshie8
  • ganeshie8
|dw:1434819455607:dw|
ParthKohli
  • ParthKohli
close enough
ganeshie8
  • ganeshie8
your painting looks way prettier than the original smith !
ParthKohli
  • ParthKohli
Of course it does - where do you think Smith was inspired from?
ganeshie8
  • ganeshie8
|dw:1434819600999:dw|
ganeshie8
  • ganeshie8
i remember something like matrix was inspired from Gita
ParthKohli
  • ParthKohli
Yes, and that is obviously not true...
mathmath333
  • mathmath333
if i have to plot the graph for this \(y\geq |1|\) then will it also be same as \(|y|\geq 1\)
hartnn
  • hartnn
|1| is just 1, right ?
mathmath333
  • mathmath333
but |-1| is also 1
hartnn
  • hartnn
but your question does not have |-1|
hartnn
  • hartnn
y >=|1| is same as y>=1 all y values greater than 1. so y will take only positive values, not negative
mathmath333
  • mathmath333
solving \(y>=|1|\)this by algebra will give me this this two inequalities \( y>=1\) and \(y>=-1\) right ?
ganeshie8
  • ganeshie8
It might help to think of \(|y|\) as \(|y-0|\), the distance between a point \(y\) on number line.and origin \(0\)
ganeshie8
  • ganeshie8
\(|1|\) is same as \(|1-0|\) the distance between points \(1\) and \(0\) on number line
hartnn
  • hartnn
if there is a variable inside |...| only then it would lead to 2 posibilities if there is a constant, then its a unique value...thats why its called a constant
mathmath333
  • mathmath333
ohk
ganeshie8
  • ganeshie8
@ParthKohli https://www.youtube.com/watch?v=re7nmrDbAG4
mathmath333
  • mathmath333
i got a question based on this one If \(|y|\geq 1\) and \(x=-|a|y\) , then which one of the following is true? \(\large \color{black}{\begin{align} a.)\ a-xy<0\hspace{.33em}\\~\\ b.)\ a-xy\geq 0\hspace{.33em}\\~\\ c.)\ a-xy>0\hspace{.33em}\\~\\ d.)\ a-xy\leq 0\hspace{.33em}\\~\\ \end{align}}\)
ParthKohli
  • ParthKohli
Since all of the questions involve the expression \(a - xy\), let's try to check its nature.\[a - xy\]\[=a - |a|y^2 \]Now \(y^2 \ge 1 \) so this expression is \(0 \) for \(y = \pm 1\) and positive \(a\). As soon as you change \(y\), it would become negative.
hartnn
  • hartnn
** a + |a|y^2
ParthKohli
  • ParthKohli
Oh wow, skipped that negative sign. =_=
ParthKohli
  • ParthKohli
OK thanks, then it should be positive. Looks good.
hartnn
  • hartnn
|a|y^2 is always positive
hartnn
  • hartnn
yes, just because |y| >=1 we can say a will be less than or = |a|y^2 numerically and that would make a+ |a|y^2 always positive :)
ParthKohli
  • ParthKohli
OK, a better word would be "nonnegative".
hartnn
  • hartnn
can be 0 ? y can't be 0
hartnn
  • hartnn
a =0 :P
mathmath333
  • mathmath333
so only options 'b' and 'd' are valid
ParthKohli
  • ParthKohli
Ew, what am I even talking about
hartnn
  • hartnn
b says a-xy is 0 or positive (non-negative) d says a-xy is 0 or negative (non-positive)
mathmath333
  • mathmath333
can \(a\) be 0
hartnn
  • hartnn
sure
mathmath333
  • mathmath333
so option be would be correct
hartnn
  • hartnn
\(\Huge \checkmark \)
mathmath333
  • mathmath333
thnx!
hartnn
  • hartnn
wlcmx!

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