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Aliun [14]
3 years ago
5

if you start from 0, how many steps do you have to take to get to -3. what does the number of steps represent​

Mathematics
2 answers:
Leviafan [203]3 years ago
5 0

Answer:

3 steps

Step-by-step explanation:

FIrst of all, if you think about it, we can just say -3 as 3 ( even though they are not equal, were just counting steps.)

On a number line, negative numbers are on the left of 0 and positive on the right of 0.

So, lets say the middle number is 0 on our number line. -3 is 3 steps to the left. We take 1 step to negative 1, another step to -2, and then a 3rd and final step to -3.

The steps represent how far the number is from 0. FIr example, we have to add 3 to -3 to get to 0, like how we add 7 to 0 to get 7!

Hope this helps!

Karo-lina-s [1.5K]3 years ago
3 0

Answer:

If you start at 0, you would have to take 3 steps back or -3 steps.

The number of steps represent the positive and negative numbers on the number line.

Step-by-step explanation:

I did the question

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<span>There are several ways to do this problem. One of them is to realize that there's only 14 possible calendars for any year (a year may start on any of 7 days, and a year may be either a leap year, or a non-leap year. So 7*2 = 14 possible calendars for any year). And since there's only 14 different possibilities, it's quite easy to perform an exhaustive search to prove that any year has between 1 and 3 Friday the 13ths. Let's first deal with non-leap years. Initially, I'll determine what day of the week the 13th falls for each month for a year that starts on Sunday. Jan - Friday Feb - Monday Mar - Monday Apr - Thursday May - Saturday Jun - Tuesday Jul - Thursday Aug - Sunday Sep - Wednesday Oct - Friday Nov - Monday Dec - Wednesday Now let's count how many times for each weekday, the 13th falls there. Sunday - 1 Monday - 3 Tuesday - 1 Wednesday - 2 Thursday - 2 Friday - 2 Saturday - 1 The key thing to notice is that there is that the number of times the 13th falls upon a weekday is always in the range of 1 to 3 days. And if the non-leap year were to start on any other day of the week, the numbers would simply rotate to the next days. The above list is generated for a year where January 1st falls on a Sunday. If instead it were to fall on a Monday, then the value above for Sunday would be the value for Monday. The value above for Monday would be the value for Tuesday, etc. So we've handled all possible non-leap years. Let's do that again for a leap year starting on a Sunday. We get: Jan - Friday Feb - Monday Mar - Tuesday Apr - Friday May - Sunday Jun - Wednesday Jul - Friday Aug - Monday Sep - Thursday Oct - Saturday Nov - Tuesday Dec - Thursday And the weekday totals are: Sunday - 1 Monday - 2 Tuesday - 2 Wednesday - 1 Thursday - 2 Friday - 3 Saturday - 1 And once again, for every weekday, the total is between 1 and 3. And the same argument applies for every leap year. And since we've covered both leap and non-leap years. Then we've demonstrated that for every possible year, Friday the 13th will happen at least once, and no more than 3 times.</span>
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tatiyna

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

We have that:

Both decomposed the fraction \frac{13}{8} into a sum of fractions. Whoever's decomposition sum has a value of \frac{13}{8} is correct.

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1 + \frac{1}{8} + \frac{3}{8} = \frac{8}{8} + \frac{1}{8} + \frac{3}{8} = \frac{8+1+3}{8} = \frac{12}{8}

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So the answer is that both are wrong.

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