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djverab [1.8K]
2 years ago
5

Five friends are leaving a bar for home. Aside from the owner of the car, one other person is deemed able to operate the vehicle

. In how many ways can these five friends be seated for the ride home?
Mathematics
1 answer:
erastova [34]2 years ago
3 0

Answer:

4

Step-by-step explanation:

one person is deemed to drive and that person is fixed on the driver's seat no mater which arrangement.

we have now 2 more seats one adjacent to the driver and one rear (two combined).

so the total ways in which all five can be arranged is as follows.

driver, adjacent to him(1)  and three back.

driver adjacent to him (different person) and three back.

see the driver is always fixed so we can ignore him.

thus we when driver set fixed , on two remaining seats (adjacent to driver and the back )there can be 4 different combinations.

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—— ——
blsea [12.9K]

Given: NQ = NT , QS Bisect NT(∴ NS=ST ) , TV Bisects QN (∴ NV=VQ )

To Prove: QS=TV

Proof: In ΔNQT

NQ=NT

\frac{1}{2}NQ=\frac{1}{2}NT

∴ VQ=ST

In a isosceles triangle, If two sides are equal then their opposites angles are equal.

∴ ∠NQT=∠NTQ       ( ∵ NQ=NT)

In ΔQST and TVQ

ST=VQ                    (sides of isosceles triangle)

∠NQT=∠NTQ          (Prove above)

QT=TQ                    (Common)

So, ΔQST ≅ TVQ  by SAS congruence property

∴ QS=TV  (CPCT)

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