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3 years ago

Distribuative property 90+27

Mathematics

2 answers:

vladimir2022 [97]3 years ago

7 0

This photo is a step by step explanation, hope it helps :)

d1i1m1o1n [39]3 years ago

6 0

Answer:

9(10 + 3)

Step-by-step explanation:

If you want to change this back into an unsolved distributive problem, find a number both 90 and 27 can be divided by, such as 9. 90/9 = 10, and 27/9 = 3. So, this would be your answer: 9(10 + 3). Hope this helps!

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3 years ago

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lisabon 2012 [21]

Answer:

B.15

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144 + 81 = 225

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4 years ago

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4 years ago

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3 years ago

Find the number of positive integers less than 100,000 whose digits are among 1, 2, 3, and 4. Hint consider the

Verdich [7]

Answer:

1364 is the number of possibilities for positive integers less than 1,00,000.

Step-by-step explanation:

1. 5 digit numbers:

We have 5 places here, and each place can have 4 options because repetition is allowed.

So, total number of possibilities for 5 digit numbers:

$4 \times 4 \times 4 \times 4 \times 4\\ \Rightarrow 1024$

2. 4 digit numbers:

We have 4 places here, and each place can have 4 options because repetition is allowed.

So, total number of possibilities for 4 digit numbers:

$4 \times 4 \times 4 \times 4 \\ \Rightarrow 256$

3. 3 digit numbers:

We have 3 places here, and each place can have 4 options because repetition is allowed.

So, total number of possibilities for 3 digit numbers:

$4 \times 4 \times 4 \\ \Rightarrow 64$

4. 2 digit numbers:

We have 2 places here, and each place can have 4 options because repetition is allowed.

So, total number of possibilities for 2 digit numbers:

$4 \times 4 \\ \Rightarrow 16$

5. 1 digit numbers:

We have 1 place here, and each place can have 4 options because repetition is allowed.

So, total number of possibilities for 1 digit numbers:

$4$

We can add all the above possibilities to find the total.

So, number of positive integers less than 100,000 whose digits are among 1, 2, 3, and 4 = 1024 + 256 + 64 + 16 + 4 = 1364

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3 years ago