Idk sorry I am not that good at math :)
Answer:
The three digit number = 951
Step-by-step explanation:
Let suppose the numbers are:
= abc
According to given condition:
a+ b + c = 15 -------------eq1
Also, given the difference between the first two digit = the difference between the last two digits:
==> l a-b l = l b-cl
==> (a-b) = (b-c)
==> (a+c) = 2b
Now we will substitue in eq1
==> a+ b + c = 15
==> 2b + b = 15
==> 3b = 15
Dividing both sides by 3 we get:
b =5
a + c = 2b
a+ c = 10
a = 10 -c ..........(2)
We know that"
(a-b) = (b-c)
==> a > b+c
==> a > 5 + c
==> 10 -c > 5 +c
==>5 > 2c
==> 2.5 > c
As c is an odd number so c will be equal to 1
c = 1
a = 10 -1
a = 9
The three digit number = 951
The hundred digit is greater than the sum of the tens and ones digits
i hope it will help you!
19 percent.
15/80 = <span>18.75, rounded is 19.</span>
The wording of this question is a bit confusing... You can't write a sequence in sigma notation, but rather a series or sum. I think the question is asking you to write the sum of the sequence,
which would be
in sigma notation.
To do this, notice that the denominator in each term is a power of 2, starting with 
and ending with 
. So in sigma notation, this series is
