Can you help me solve this??

1.solve s= -1/2gt^2+vt

No clue.

2.

Two cars leave City A at the same time to travel

to City B. One car travels at 72km/h and the other

at 78km/h. If the slower car arrives 20 min after

the faster car, how far apart are City A and City B?

Note that 20 minutes = 1/3 of an hour

T=time for faster car

T + 1/3 = time for slower car

78k/h * T = Distance between cities

72k/h * (T + 1/3) = Distance between cities

78*T = 72 * (T + 1/3)

78T = 72T + 78/3

6T = 26

T = 26/6

T = 13/3 = 4 + 1/3 hours

The cities are 78*(13/3) = 338 miles apart.

As a back check:

338/78 = 4.33 hours

338/72 = 4.69 hours

which differ by approximately 1/3 of an hour.

3. Determine K so that the line through (k+2,k) and (3,2) has a slope of 3.

The slope-intercept form is:

y=mx+b

Plug in the two points that we're given:

(y-2) = 3(x-3) + B

(y-K) = 3(x-(K+2)) + B

Solve for B in each case:

(y-2) - 3(x-3) = B

(y-K) - 3(x-(K+2)) = B

Since both expressions equal B, solve for K:

(y-2) - 3(x-3) = (y-K) -3(x - (K+2))

y-2-3x+9 = y-K-3(x-K-2)

y-2-3x+9 = y-K-3x+3K+6

7-3x = 2K-3x+6

7 = 2K+6

1 = 2K

1/2 = K

The the two points are (2.5,0.5) and (3,2). The slope

is rise/run, which is (2-0.5)/(3-2.5) = 1.5/.5 = 3.0

4. Solve the system: 3c-d=1 and 3d+c=1

Solving the second equation for "c":

c = 1-3d

Substituting into the first equation:

3(c)-d=1

3(1-3d)-d=1

3-9d-d=1

3-10d=1

2=10d

1/5 = d

c = 1-3d,

c = 1-3/5

c = 2/5

To check:

3c - d = 1

3(2/5) - (1/5) = 1

6/5 - 1/5 = 1

and

3d+c = 1

3(1/5) + 2/5 = 1

3/5 + 2/5 = 1

OK.