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OpenStudy (anonymous):
What's the question?
OpenStudy (anonymous):
y=mx+b^2
y=m
thats it
OpenStudy (anonymous):
sorry but we dont understand that
OpenStudy (amistre64):
b is a constant; so it doesnt matter what degree it is
OpenStudy (anonymous):
You have two lines, are we to figure out their intersection? What are we solving?
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OpenStudy (amistre64):
its 16...no it isnt it 4^2... no it isnt its sqrt(256)...no it isnt it 9+7.....
OpenStudy (anonymous):
thats all my book says its says i dont understand it its so confusing
sove these problems with a graph
y=mx+b^2
y=b
OpenStudy (anonymous):
ugh i give up.
OpenStudy (anonymous):
Don't give up, that second statement that you typed, y=b. The first time it was y=m which one is it?
OpenStudy (anonymous):
sorry i ment y=m
y=m
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OpenStudy (anonymous):
So it says to solve by graphing, y=m (assuming m is just a number and not the slope) is just a horizontal line...as for the y=mx+b^2 I don't know how to graph that without points to help me find the b (y-intercept)
OpenStudy (anonymous):
if y=m and m is the slope then you have a line that starts at (0,0) and has a positive slope of 1
OpenStudy (anonymous):
but then it should say y=mx
OpenStudy (anonymous):
or y=x sorry...
OpenStudy (anonymous):
wait...THATS IT! IT DOSE START AT 0,0 THANKS SOOO MUCH!!!!!!!
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