2cars started to move at the same time, at the same direction but one was moving twice as fast as the other. 6 hours later, the 2 cars were 204 miles apart. find the speed for each car.

The speeds are 34 mph for the slower car and 68 mph for the faster car.

Step-by-step explanation:

speed = distance/time

Using s for speed, d for distance, and t for time, we have the equation for speed:

s = d/t

Solve for distance, d, by multiplying both sides by t.

d = st

Now we use the given information.

Speed of slower car: s

Speed of faster car: 2s

Distance traveled by faster car: d

Distance traveled by slower car: d - 204

time traveled by faster car = time traveled by slower car = 6

Distance equation for faster car:

d = st

d = 2s * 6

d = 12s     < equation 1

Distance equation for slower car:

d = st

d - 204 = s * 6

d - 204 = 6s

d = 6s + 204    < equation 2

Now, using equations 1 and 2, we have a system of two equations in two unknowns.

d = 12s

d = 6s + 204

Since the first equation is already solved for d, we can use the substitution method. Substitute 12s for d in the second equation:

12s = 6s + 204

6s = 204

s = 34

The speed of the slower car is 34 mph.

The speed of the faster car is

2s = 2(34) = 68

The speed of the faster care is 68 mph.

answer. d.289 hope that

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)