The braking distance dv (in meters) for a certain car moving at velocity v (in meters/second) is given by =dvv234.
the car's velocity bt (in meters/second) t seconds after starting is given by =bt3t.

write a formula for the braking distance st (in meters) after t seconds.

it is not necessary to simplify.

step-by-step explanation:

the answer is 128

a 90 degree rotation across the the origin followed by a reflection across the x-axis.

step-by-step explanation:

35/2 is it i think maybe

part a: triangles gcb and epb

part b. sides cg and rp are parallel by definition of parallelogram.

angles cgb and e are congruent by being alternate interior angles of parallel lines cut by a transversal. angles gbc and ebp are congruent because they are vertical angles. by aa similarity, the triangles are similar.

part c. since the triangles are similar, the ratios of the lengths of corresponding sides are equal.

calculation of be:

gb/be = cb/bp

(400 ft)/be = 300/200

(400 ft)/be = 3/2

3be = 2 * 400 ft

3be = 800 ft

be = 800/3 ft

calculation of pe:

gc/pe = cb/bp

(350 ft)/pe = 300/200

(350 ft)/pe = 3/2

3pe = 2 * 350 ft

3pe = 700 ft

pe = 700/3 ft