Write an equation in point-slope form for the line through the given point with the given slope. (8, 3); m = 6
what's the point slope form? \[y-y_1=m(x-x_1)?\]
lol
Yes it is apparently: so you have your point (8;3), so x1=8 and y1=3, and m=6. Thus: \[y-3=6(x-8)\]
@dumbcow why lol?
: Suppose y varies directly with x, and y = 8 when x = –6. What direct variation equation relates x and y? What is the value of y when x = –2?
Get it in y=mx+b form first.
or is this a diff question?
y=8+-6x
" Suppose y varies directly with x, and y = 8 when x = –6. What direct variation equation relates x and y? What is the value of y when x = –2?" Is this a separate question from your original one?
yea
.....
I had to look this up.
-6/8 = -2/y This is a direct variation. Cross multiply and solve for y.
-6y = -16
its good explaining 0/375
0.375
Nikita, would you prefer just a numerical answer? That will help so much more for your exams.
yup
Believe me none of us remembers every single little thing from algebra. Lol I had to google direct variation of a line.
Suppose y varies directly with x, and y = 8 when x = –6. What direct variation equation relates x and y? What is the value of y when x = –2? A. y = –0.75x; 1.50 B. y = –1.33x; 2.67 C. y = 1.33x; –2.67 D. y = 0.13x; –0.25
Hmm... still thinking
y = k*x Solve for k. 8 = k *(-6) k = -8/6 = -4/3 y = (-4/3)*x Plug in x=-2 into x and solve. y=(-4/3)*(-2) y=8/3 I dunno, I think this is wrong.
b/c y=2.67 this way :-/
Just pick A. B/c the method is absolutely correct. Someone *'ed up on writing the answer choices.
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