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lisabon 2012 [21]
2 years ago
11

When I get multiplied by any number, the sum of the figures in the product is always me. What am I?

Mathematics
2 answers:
vladimir1956 [14]2 years ago
8 0

Answer: 9

Step-by-step explanation:

When you multiply 9*9, you get 81 whose sum is 9 (8+1)

Likewise, 9*8=72( 7+2=9)

Andre45 [30]2 years ago
8 0

I am the number '9'

To prove:

On multiplying 9 with different numbers, the sum of the digits of its product is always 9. Let's see;

9 * 2 = 18

where (1 + 8 = 9)

Now, multiplying it by 3

9 * 3 = 27

where (2 + 7 = 9)

Similarly,

9 * 4 = 36

where (3 + 6 = 9)

9 * 5 = 45

where (4 + 5 = 9)

9 * 6 = 54

where (5 + 4 = 9)

9 * 7 = 63

where (6 + 3 = 9).

Thus, option b i.e. 9 is the correct answer.

Learn more about 'numbers' here:

brainly.in/question/24738334

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Quadrilateral WXYZ is dilated by a scale factor of į to form quadrilateral W'X'Y'Z'. What is the measure of side ZW? Z 28 Y" 42
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Answer:

See Explanation

Step-by-step explanation:

Given

<em>The given side lengths are not clear enough:</em>

<em>So, I will give a general explanation.</em>

<em />

From the question, we understand WXYZ is dilated to produce W'X'Y'Z

This implies that the sides of WYZY is multiplied by the scale factor (k) to give W'X'Y'Z'

i.e.

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Divide both sides by k to get side length ZW

ZW = \frac{Z'W'}{k}

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<em>Apply these steps and you will get your answer</em>

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