Dwayne drove 18 miles to the airport to pick up his father and then returned home. On the return trip he was able to drive an average of 15 miles per hour faster than he did on the trip there. If the total driving time was 1 hour, then what was his average speed driving to the airport? Someone please help

30 mph

Step-by-step explanation:

Let d = distance (in miles)

Let t = time (in hours)

Let v = average speed driving to the airport (in mph)

⇒ v + 15 = average speed driving from the airport (in mph)

Using:  distance = speed x time

Create two equations for the journey to and from the airport, given that the distance one way is 18 miles:

We are told that the total driving time is 1 hour, so the sum of these expressions equals 1 hour:

Now all we have to do is solve the equation for v:

As v is positive, v = 30 only

So the average speed driving to the airport was 30 mph

(and the average speed driving from the airport was 45 mph)

SI=(P*R*T)/100

P=2000

R=1.5

T=6

SI=(2000*1.5*6)/100

=(2000*9)/100

=180

Neil will earn interest of 180

The answer is in the image 

The total nom of  code that can be used is equal to 5+3 = 8

The solution is in the following image

y=2x+15

where y=Value of coin

x=Age in years

Value of coin after 19 years=2*19+15

=$53

Therefore, Value after 19 years=$53

F=ma

where F=force

m=mass

a=acceleration

Here,

F=4300

a=3.3m/s2

m=F/a

    =4300/3.3

    =1303.03kg

F=ma

where F=force

m=mass

a=acceleration

Here,

F=4300

a=3.3m/s2

m=F/a

    =4300/3.3

    =1303.03kg

Approximately it is aqual to 1300kg

The wood before starting =12 feet

Left wood=6 feet

Wood used till now=12-6=6 feet

Picture frame built till now= 6/(3/4)

=8 pieces

Therefore, till now 8 pieces have been made.