Write three numbers for which you would use the hundreds place to compare to 35,712

If we want to use a particular digit place to compare two numbers, then:

all the digits at the left of that place must be equal in both numbers.

So, for example, if we want to use the hundreds place to compare the number 35,712, we will only do it for numbers like:

35,231

35,567

35,887

35,273

Where "compare" in this case means that we take the difference between the hundreds places:

using 35,712 and 35,231 that difference is just:

7 - 2 = 5

this means that 35,712 is roughly 500 more than 35,712

Why do we have the rule?

Because if any of the numbers at the left were different, then wouldn't make any sense to compare the hundreds places because we would have a difference of thousands or tens of thousands which are significantly larger than a difference in hundreds.

And for that reason is that the numbers at the right don't matter, because if we have a difference of hundreds, we can ignore the differences of tens or units, as those are a lot smaller.

If you want to learn more, you can read:

link

Step-by-step explanation:

If you're using the hundreds place for comparison, it means the digits in higher places are the same, and that digits in the hundreds place are different.

This will be true for any 5-digit number of the form ...

  (3)(5)(not 7)(something)(something).

The numbers might have digits to the right of the decimal point, but those are irrelevant, as are the digits in the 10s and 1s places.

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!