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2 years ago

Find the sum of all the digits used to write the numbers from 0 to 9999

Mathematics

2 answers:

stealth61 [152]2 years ago

8 0

Answer: 49,995,000

Step-by-step explanation:

0 + 9999 = 9999

1 + 9998 = 9999

0 to 9999 there would be 5000 pairs

5000 x 9999 = 49995000

zhenek [66]2 years ago

3 0

38,890 is the right answer
Hope it helped

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2 years ago

Using properties of sets show that : a) A ∩ (A’ U B) = A ∩ B b) A ∩ (A U B )’ = Ф

asambeis [7]

Answer:

a) From A ∩ A' = ∅, we have;

A ∩ (A' ∪ B) = A ∩ B

b) From A ∩ (A' ∩ B') = (A ∩ A') ∩ B' and A ∩ A' = ∅, we have;

A ∩ (A ∪ B)' = ∅

Step-by-step explanation:

a) By distributive law of sets, we have;

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

From the complementary law of sets, we have;

A ∩ A' = ∅

Therefore, for A ∩ (A' ∪ B) = A ∩ B, we have

A ∩ (A' ∪ B) = (A ∩ A') ∪ (A ∩ B) (distributive law of sets)

A ∩ A' = ∅ (complementary law of sets)

Therefore;

(A ∩ A') ∪ (A ∩ B) = ∅ ∪ (A ∩ B) = (A ∩ B) (Addition to zero identity property)

∴ A ∩ (A' ∪ B) = A ∩ B

b) By De Morgan's law

(A ∪ B)' = A' ∩ B'

Therefore, A ∩ (A ∪ B)' = A ∩ (A' ∩ B')

By associative law of sets, we have;

A ∩ (A' ∩ B') = (A ∩ A') ∩ B'

A ∩ A' = ∅ (complementary law of sets)

Therefore, (A ∩ A') ∩ B' = ∅ ∩ B' = ∅

Which gives;

A ∩ (A ∪ B)' = ∅.

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