Which is an equation in point-slope form for the given point and slope? Point: (5, 9); Slope: 2 A. y – 5 = 2(x + 9) B. y + 9 = 2(x – 5) C. y – 9 = 2(x + 5) D. y – 9 = 2(x – 5)
point slope form: y - y1 = m(x - x1) when m = slope and given the point (x1, y1) so when given Point: (5, 9); Slope: 2 plug in m = 2 x1 = 5 and y1 = 9 into y - y1 = m(x - x1)
d.
yep :)
thx I got 5 more questions though
I can help with those too :)
Which is an equation in point-slope form for the given point and slope? Point: (–8, 3); Slope: 6 A. y + 3 = 6x – 48 B. y + 3 = 6(x + 8) C. y – 3 = 6(x – 8) D. y – 3 = 6(x + 8)
same idea as before :) point slope form: y - y1 = m(x - x1) when m = slope and given the point (x1, y1) so when given Point: (–8, 3); Slope: 6 plug in m = 6 x1 = -8 and y1 = 3 into y - y1 = m(x - x1)
c.
be careful :) when you plug in the x1 = -8 y - y1 = m(x - (-8)) ^ negative times negative equals positive
ohhh b.
the y1 = 3 would remain negative so you would have gotten y - 3 = 6(x + 8)
d. definitely d.
yep :)
thx :D
you're welcome :)
2 more instead of 3
sure :)
Which is an equation in point-slope form for the given point and slope? Point: (1, –7); Slope: \[-\frac{ 2 }{3} \]
same idea as before :) point slope form: y - y1 = m(x - x1) when m = slope and given the point (x1, y1) so when given Point: (1, –7); Slope: −2/3 plug in m = -2/3 x1 = 1 and y1 = -7 into y - y1 = m(x - x1)
\[y-7= -\frac{2}{3}\]
\[y+7=\frac{ 2 }{ 3 }\]
\[y-7=\frac{ 2 }{ 3 }\]
\[y+7=-\frac{ 2 }{ 3 }\]
\[y - (-7) = (-\frac{ 2 }{ 3 }) (x - (1))\]
d.
yep :)
:D one last question
ok :)
A yearbook printer charges based on the number of pages printed. Here is a table that shows the cost of some recent yearbooks:
Which equation gives the cost, y, in terms of the number of pages, x? A. y = 83.3x + 0.65 B. y = 0.012x + 49.985 C. y = 0.012x + 0.65 D. y = 0.012x + 1.25
1.85 minus 1.25 = 0.6 so every 50 pages = 0.6 then 0.6 divided by 50 = how much one page costs = 0.012 when pages = 0 cost = 1.25 - 0.6 = 0.65 so when y = mx + b m = 0.012 and b = 0.65
c
yep :)
thx! :D
you're welcome :)
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