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Ipatiy [6.2K]
2 years ago
14

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?
Mathematics
1 answer:
Tju [1.3M]2 years ago
5 0

The houses can be placed in 362,880 ways.

<u>Step-by-step explanation:</u>

The 9 houses are each in different design.

The each lot can place any of the 9 houses.

  • The 1st lot can place anyone house of all the 9 houses.
  • The 2nd lot can place one of remaining 8 houses.
  • The 3rd lot can place one of remaining 7 houses.

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

<u>From the above steps, it can be determined that :</u>

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.

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Which statement identifies how to show that j(x) = 11.6ex and k(x) = In (StartFraction x Over 11.6 EndFraction) are inverse func
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<h2>It must be shown that both j(k(x)) and k(j(x)) equal x</h2>

Step-by-step explanation:

Given the function  j(x) = 11.6e^x and k(x) = ln \dfrac{x}{11.6}, to show that both equality functions are true, all we need to show is that both  j(k(x)) and k(j(x)) equal x,

For j(k(x));

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j[(ln x/11.6)] = 11.6 * x/11.6

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