What is the equation of the graphed linear model?
y = ___ x +___
http://static.k12.com/calms_media/media/1549000_1549500/1549109/2/a4d8cb5972366f402912ed1759aeadaa5560d1f1/MS_IMC-141014-181213.jpg
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OpenStudy (anonymous):
@Preetha can you help me??
OpenStudy (anonymous):
@amistre64
OpenStudy (anonymous):
hia again
OpenStudy (amistre64):
whats our slope? and hi
OpenStudy (anonymous):
idk yet because idk what numbers to use
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OpenStudy (anonymous):
y=mx+b
OpenStudy (amistre64):
the line goes thru 2 readily seen points. use them
OpenStudy (amistre64):
what are the points?
OpenStudy (anonymous):
oh (1,3) and (10,11)
OpenStudy (amistre64):
9,11 but yeah
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OpenStudy (anonymous):
oh ya lol
OpenStudy (anonymous):
so i have to do Y1 - Y2
------
X1 - X2
OpenStudy (amistre64):
you prolly could yeah :)
OpenStudy (anonymous):
is the slope 1
OpenStudy (amistre64):
yep, and when x=0 what would you gander the y intercept is from the chart?
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OpenStudy (anonymous):
uh 0 i think so b= 0
OpenStudy (anonymous):
so y=mx+b = y= 1 x + b
OpenStudy (anonymous):
y=1x+0
OpenStudy (amistre64):
the line is not crossing the y axis anywhere near y=0 ...
we can do this the hard way if you want ..
given a point (a,b) and a slope, m: we can construct the "point-slope" form if a line as:
y-b = m(x-a) and then conform it to your particular setup as
y = mx -ma + b
keeping in mind that this b is not your b
OpenStudy (amistre64):
or we could just adjust it by the point
1,3 is on the line sooo
y = mx + b
3 = 1(1) + b ; what is b?
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OpenStudy (anonymous):
i think 1
OpenStudy (anonymous):
I forgot how to find the y-intercept
OpenStudy (amistre64):
3 = 1 + 1 ... not quite right
OpenStudy (anonymous):
oh 2 lol
OpenStudy (anonymous):
maybe
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OpenStudy (amistre64):
you find the y intercept (the point it cross the y axis) by looking at the chart or filling in the values
OpenStudy (amistre64):
3 = 1 + b, so b = 2 yes
there are many ways to approach this.
OpenStudy (anonymous):
ya i feel dumb
OpenStudy (anonymous):
thank you again theirs your medal
OpenStudy (amistre64):
good luck :)
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OpenStudy (anonymous):
do you think you can help me with this one or check it