Unlike the previous problem, this one requires application of the Law of Cosines. You want to find angle Q when you know the lengths of all 3 sides of the triangle.
Law of Cosines: a^2 = b^2 + c^2 - 2bc cos A
Applying that here:
40^2 = 32^2 + 64^2 - 2(32)(64)cos Q
Do the math. Solve for cos Q, and then find Q in degrees and Q in radians.
(-3,2) (0,0) i think. it’s been a whole
Answer:
The line segment that is a radius is FD.
Hope it will help:)❤
Y-y1=m(x-x1)
a point on the line is (x1,y1) and the slope is m
so
slope between (x1,y1) and (x2,y2) is (y2-y1)/(x2-x1)
so
points (-2,3) and (3,0)
slope is (0-3)/(3-(-2))=-3/(3+2)=-3/5
a point is (3,0) Or (-2,3)
so it could be
y-0=-3/5(x-3) or y-3=-3/5(x-(-2)) which is equal to y-3=-3/5(x+2)
that's the answer
B is answer
Answer:
-14-x
Step-by-step explanation:
Distribute the negative.