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Mathematics 8 Online
OpenStudy (anonymous):

NEED HELP! MEDAL! Which equation is shown on the graph? A. y = 2x + 2 B. y = x + 2 C. y = 2x D. y = x – 2

OpenStudy (rarepepe):

whats the graph look like?

OpenStudy (anonymous):

@AlexandervonHumboldt2

OpenStudy (rarepepe):

c

OpenStudy (rarepepe):

I mean b sorry

OpenStudy (anonymous):

explain plz.

OpenStudy (alexandervonhumboldt2):

ok so first count the slope on the grath: chose points (-2, 0) (0, 2)

OpenStudy (anonymous):

uh hu and then..

OpenStudy (alexandervonhumboldt2):

use slope formula (y1-y2)/(x1-x2)=(0-2)/(-2-0)=-2/-2=1

OpenStudy (anonymous):

?

OpenStudy (alexandervonhumboldt2):

so the slope is 1

OpenStudy (anonymous):

ok i still don't get it

OpenStudy (alexandervonhumboldt2):

the grath is 2 units upper than the 0, 0 point. thus b in the equation y=x+b is 2 thus the equation is t=x+2

OpenStudy (alexandervonhumboldt2):

woops i meant y=x+2

OpenStudy (anonymous):

i see thxs!

OpenStudy (anonymous):

can you stay to help me on other questions!

OpenStudy (anonymous):

so who is correct?

OpenStudy (anonymous):

If x can be any number, how many solutions are there for the equation? y = 4x – 1 A. There are two solutions. B. There is no solution. C. There are many solutions. D. There is only one solution.

OpenStudy (anonymous):

@texaschic101

OpenStudy (texaschic101):

I was wrong on the first one...I miscalculated...sorry. If x can be any number, and the equation still be true, then there is infinite solutions

OpenStudy (anonymous):

oh ok its ok.

OpenStudy (anonymous):

can u solve second please @texaschic101

OpenStudy (texaschic101):

already did....if x can be any number, and the equation still be true, then there is infinite solutions

OpenStudy (anonymous):

thxs!

OpenStudy (texaschic101):

glad to help :)

OpenStudy (anonymous):

The vertices of a right triangle are (0, 5), (5, 0), and (5, y). What is the value of y? ( , )

OpenStudy (anonymous):

@texaschic101

OpenStudy (texaschic101):

I am thinking (5,5)...but you might want to get a second opinion

OpenStudy (anonymous):

ok I was thinking same but wasnt sure so yeah @texaschic101

OpenStudy (anonymous):

@AlexandervonHumboldt2 please solve second question

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