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Mathematics 16 Online
OpenStudy (hypersniper445):

Can anyone help with 3 questions pls? i will fan and metal

OpenStudy (hypersniper445):

OpenStudy (hypersniper445):

I got a

OpenStudy (hypersniper445):

A and D

OpenStudy (hypersniper445):

OpenStudy (hypersniper445):

is that right

OpenStudy (hypersniper445):

hello? @HannahC234 @damirax

OpenStudy (anonymous):

what

OpenStudy (hypersniper445):

you were on my question but not saying anything

OpenStudy (hannahc234):

For the first question One way that I just made up is to try to convert the equations to slope intercept form and see if it is in normal form, if it is in y=mx+b form, then the equation is linear, if the equation cannot be converted, then it is not linear, in the y=mx+b form, the exponent above the x and y must be 1 you could also convert them to conic section equaitons and see what shape the graph is so we do 1. y-4=-8(x-1) distribute y-4=-8x+8 add 4 to both sides y=-8x+12 this is in y=mx+b form so it is linear 2. y=x^4 this is not linear, it is an exponential function graph and is a curve 3. y-x^2=4.5 add x^2 to both sides y=x^2+4.5 exponent is not =1 so this is not linear 4. 3x+5y=15 you should recognize that this is linear but anyway subtract 3x from both sides 5y=-3x+15 divide by 5 y=-3/5x+3 linear one way is to focus on the placeholders and their powers. if they can be both simplified to equal 1, then it is linear, answer: 1.Yes 2.No 3 No 4. Yes focus on the placeholders and their powers. if they can be both simplified to equal 1, then it is linear

OpenStudy (hypersniper445):

oh wow thank you

OpenStudy (hannahc234):

No problem

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