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uranmaximum [27]
3 years ago
7

Given 11 different natural numbers, none greater than 20. Prove that two of these can be chosen, one of which divides the other.

Mathematics
2 answers:
Thepotemich [5.8K]3 years ago
7 0

Answer:

777 838 3

Step-by-step explanation:

sood

frosja888 [35]3 years ago
4 0

Answer:11 honestly have no idea but im going to give my best answer it is 31

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Which of the following expressions could be used to find the sale price of an item that costs d dollars if it is on sale for 30%
Elenna [48]

Greetings!

An item at full price is 100%, or by using the percentage formula: (\frac{percentage}{100})

If something is at 30% off, that means it is 70% of the original price (because minus the percentage from the 100%)

So by substituting the numbers into the percentage formula you end up with:

(\frac{70}{100}) which is 0.7

So your answer would be D, 0.7!


Hope this helps!

6 0
3 years ago
What is the difference between DEPENDENT and INDEPENDENT probability?
ira [324]

Answer:

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2 years ago
Hello can u please help
Andrei [34K]

y = mx + b

-3 = -2(-5) + b

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The line would pass through -13 on the y-axis and it would go down 2 and move to the right 1 for every new point.

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6 0
2 years ago
Read 2 more answers
What is a three digit number thats an odd multiple of three and the product of its digits is 24 and is larger than 15 squared?
Reika [66]
The only way 3 digits can have product 24 is 
1 x 3 x 8 = 241 x 4 x 6 = 242 x 2 x 6 = 242 x 3 x 4 = 24
So the digits comprises of 1,3,8 or 1,4,6, or 2,2,6, or 2,3,4 
To be divisible by 3 the sum of the digits must be divisible by 3.
1+  3+ 8=12, 1+ 4+ 6= 11, 2 +2 + 6=10, 2 +3 + 4=9Of those sums of digits, only 12 and 9 are divisible by 3. 

So we have ruled out all but integers whose digits consist of1,3,8, and 2,3,4.

Meanwhile they must be odd they either must end in 1 or 3.
The only ones which can end in 1 are 381 and 831.

The others must end in 3. 

They must be greater than 152 which is 225. So the

First digit cannot be 1. So the only way its digits can contain of1,3,8 and close in 3 is to be 813.

The rest must contain of the digits 2,3,4, and the only way they can end in 3 is to be 243 or 423.  
So there are precisely five such three-digit integers: 381, 831, 813, 243, and 423.
7 0
3 years ago
the width of a rectangle is 1 units less than the length. the area of the rectangle is 56 units. what is the width, in units, of
Oksana_A [137]
Answer: Length = 8 units, Width = 7 units
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3 years ago
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