The slope intercept form using the two points would be : y= 2x-13
let's bear in mind that sin(θ) in this case is positive, that happens only in the I and II Quadrants, where the cosine/adjacent are positive and negative respectively.
Gf(x)
=g[f(x)]
=g[x2-1]
=x2-1+2
Answer:
A. (x−2y)(y−3x)
=(x+−2y)(y+−3x)
=(x)(y)+(x)(−3x)+(−2y)(y)+(−2y)(−3x)
=xy−3x2−2y2+6xy
=−3x2+7xy−2y2
B. (2p+3)(p2−4p−7)
=(2p+3)(p2+−4p+−7)
=(2p)(p2)+(2p)(−4p)+(2p)(−7)+(3)(p2)+(3)(−4p)+(3)(−7)
=2p3−8p2−14p+3p2−12p−21
=2p3−5p2−26p−21
Hope it helps
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The equation is y = x + 2 and they are perpendicular lines.
A rotation of 270 degrees means that you turn 3/4 of the way around a circle. That would put the y-intercept at (0, 2) and the line would be going up instead of down.