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3 years ago

How many different 2-digit numbers are there with the following property: the tenth digit is greater than the units digit?

Mathematics

1 answer:

EastWind [94]3 years ago

8 0

Answer:

There is 45 different 2-digit numbers.

Step-by-step explanation:

My way is kinda dumb, but it still works. So, 2 digit numbers is from 10-99. We can start from the 10-19 first.

10, 11,12,13,14,15,16,17,18,19

We need to find the numbers that the tenth digit is greater than the units digit.

10, 11,12,13,14,15,16,17,18,19

Since 1 is the tenth digit, all the ones digits are all going to be bigger. Same goes with 20-29. Then you will have 2 numbers that the tenth digit is greater. My way applies when the tenth digit is only less than the ones digit, not greater. If you do this way...

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

10-19: 1 number

20-29: 2 numbers

30-39: 3 numbers

40-49: 4 numbers

50-59: 5 numbers

60-69: 6 numbers

70-79: 7 numbers

80-89: 8 numbers

90-99: 9 numbers

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

So, if you add 1+2+3+4+....+9(which all of you probably memorized by now) it would be 45. The answer is 45. Hope this helped!

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2 years ago

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Answer:

Step-by-step explanation:

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2 years ago

A leprechaun places a magic penny under a girls pillow. The next night there are 2 magic pennies under her pillow. The following

kiruha [24]

Answer:

After 47 days she will have more than 90 trillion pennies.

Step-by-step explanation:

At the beginning there was 1 penny. At the second day the amount of pennies under the pillow became 2.

The amount of pennies doubled each day. So the series is,

$1,2,4,8,16,32,.....$

This series is in geometric progression.

As the pennies from each of the previous days are not being stored away until more pennies magically appear so the sum of series will be,

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where,

a = initial term = 1,

r = common ratio = 2,

As we have find the number of days that would elapse before she has a total of more than 90 trillion, so

$\Rightarrow 90\times 10^{12}\le \dfrac{1(2^n-1)}{2-1}$

$\Rightarrow 90\times 10^{12}\le \dfrac{2^n-1}{1}$

$\Rightarrow 90\times 10^{12}\le 2^n-1$

$\Rightarrow 2^n\ge 90\times 10^{12}+1$

$\Rightarrow \log 2^n\ge \log (90\times 10^{12}+1)$

$\Rightarrow n\times \log 2\ge \log (90\times 10^{12}+1)$

$\Rightarrow n \ge \dfrac{\log (90\times 10^{12}+1)}{\log 2}$

$\Rightarrow n \ge 46.4$

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3 years ago

Find the x is qs bisects pqr and pqr=82

slavikrds [6]

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$\bf \stackrel{\measuredangle PQS}{10x+1}=41\implies 10x=40\implies x=\cfrac{40}{10}\implies x=4$

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3 years ago

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aksik [14]

Answer:

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3 years ago