Contractor wishes to build 9 houses, each different in design. In how many ways can he place these houses on a street if 6 lots are on one side of the street and 3 lots are on the opposite side?

The houses can be placed in 362,880 ways.

Step-by-step explanation:

The 9 houses are each in different design.

The each lot can place any of the 9 houses.

Similarly, the process gets repeated until the last house is placed on a lot.

From the above steps, it can be determined that :

The number of ways to place the 9 houses in 9 lots = 9!

⇒ 9×8×7×6×5×4×3×2×1

⇒ 362880 ways.

Therefore, the houses can be placed in 362880 ways.

47.28 or 47.2857143

step-by-step explanation:

you just divide 331 by 7 and you get that answer. ( i think so tho)

b. 18

step-by-step explanation:

so the equation for this is:

ab^2=ac⋅ad

ab being the top length of 6 and ac being the length of 2 and ad being the length of x. let's plug into the equation:

6^2 = 2x

now lets just solve for x:

36 = 2x

18 = x

i hope that !