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2 years ago

Titus is asked to prove hexagon FEDCBA is congruent to hexagon

Mathematics

1 answer:

Goshia [24]2 years ago

5 0

Answer:

(x, y) -------> ( - x , y - 10 )

Step-by-step explanation:

Titus is not correct.

There are two transformations will correctly prove FEDCBA ≅ F'E'D'C'B'A'

First transformation is Reflection over y-axis , then translation 10 units down.

The final formula is ( - x , y - 10 )

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3 years ago

A company rents out 13 food booths and 27 game booths at the county fair. The fee for a food booth

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Answer:

50+(7d*13)+80+(9d*27)

Step-by-step explanation:

So, there are 13 food booths and each food booth is $50 plus $7 per day.

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2 years ago

Find the number of elements in A1 ∪A2 ∪A3 if there are 100 elements in A1, 1000 in A2, and 10,000 in A3 if a) A1 ⊆ A2 and A2 ⊆ A

Stella [2.4K]

Answer:

Step-by-step explanation:

Given that there are 3 sets such that there are 100 elements in A1, 1000 in A2, and 10,000 in A3

a) If A1 ⊆ A2 and A2 ⊆ A3

then union will contain the same number of elements as that of A3

i.e. $n(A1 ∪A2 ∪A3)=n(A3) =10000$

b) If the sets are pairwise disjoint.

union will contain the sum of elements of each set

$n(A1 ∪A2 ∪A3) = 100+1000+10000=11100$

c) If there are two elements common to each pair of sets and one element in all three sets

We subtract common elements pairwise and add common element in 3

i.e. $n(A1 ∪A2 ∪A3) = 100+1000+10000-2-2-2+1\\= 10995$

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3 years ago