Two skaters are racing towards the finish line of the race. the first skater has a 40 meter lead and is traveling at a rate of 12 meter. the second skater is traveling at a rate of 14 meter per second. how long will it take for the second skater to pass the first skater?

20 seconds

Step-by-step explanation:

The distance to cover between the lead skater and trailing skater is 40 meters

The lead skater's rate is 12 m/s

The trailing one's rate is 14 m/s

So, we can say:

second skater is accelerating, or covering up the distance, in 14 - 12 = 2 m/s

So, at a rate of 2 meters per second, how long would it take to cover 40 meters??

We use the distance formula, substitute the known and solve for unknown.

Distance Formula = D = RT

Where

D is distance

R is speed/rate

T is time

We know

r = 2 m/s

d = 40 m

t = ?

Now,

D = RT

40 = 2t

t = 40/2

t = 20 (seconds)

Hence,

it will take 20 seconds for the second skater to catch up (pass)

step-by-step explanation:

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.