Given that we assume that all the bases of the triangles are parallel.
We can use AAA or Angle-Angle-Angle to prove that these triangles are similar.
Each parallel line creates the same angle when intersecting with the same side.
For example:
The bases of each triangle cross the left side of all the triangle.
Each angle made by the intersecting of the the parallel base and the side are the same.
Thus, each corresponding angle of all the triangles are congruent.
If these angles are congruent, then we have similar triangles.
Answer:
Step-by-step explanation:
Part A
xf = xo + vo* t + 1/2 a*t^2 Subtract xo
xf - xo = 0*t + 1/2 a*t^2 multiply by 2
2(xf - xo) = at^2 divide by t^2
2(xf - xo ) / t^2 = a
Part B
Givens
xo =0
vo = 0
a = 10 m/s^2
xf = 120 m
Solution
xf = xo + vo* t + 1/2 a*t^2 Substitute the givens
120 = 0 + 0 + 1/2 * 10 * t^2 Multiply by 2
120*2 = 10* t^2
240 = 10*t^2 Divide by 10
240/10 = t^2
24 = t^2 take the square root of both sides.
√24 = √t^2
t = √24
t = √(2 * 2 * 2 * 3)
t = 2√6
Answer:
ｙ = 7x - 4
just as it says in words
There’s no picture but three ordered pairs for the equation would be: (0,2/3), (1,3), (2,16/3).
