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Mathematics 13 Online
OpenStudy (anonymous):

now if y=mx+b^2 how do i find the ancwer

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?

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

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!!!!!!!

OpenStudy (anonymous):

bye

OpenStudy (anonymous):

bye :)

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