That number is the ratio of 578,778,777 to 100,000,000 , so it's rational.
Here's a surprising factoid:
ANY number that you can completely write down on paper with digits...
and a decimal point or a fraction bar if you need them ... is rational.
If you can write it down ... without saying "repeating" or using symbols ...
it's rational.
1.595 to 2 decimal points is 1.60 because the second decimal place is nearest to 10.
The length of the KN is 4.4
Step-by-step explanation:
We know from Pythagoras theorem
In a right angle ΔLMN
Base² + perpendicular² = hypotenuse
²
From the properties of triangle we also know that altitudes are ⊥ on the sides they fall.
Hence ∠LKM = ∠NKM = 90
°
Given values-
LM=12
LK=10
Let KN be “s”
⇒LN= LK + KN
⇒LN= 10+x eq 1
Coming to the Δ LKM
⇒LK²+MK²= LM²
⇒MK²= 12²-10²
⇒MK²= 44 eq 2
Now in Δ MKN
⇒MK²+ KN²= MN²
⇒44+s²= MN² eq 3
In Δ LMN
⇒LM²+MN²= LN²
Using the values of MN² and LN² from the previous equations
⇒12² + 44+s²= (10+s)
²
⇒144+44+s²= 100+s²+20s
⇒188+s²= 100+s²+20s cancelling the common term “s²”
⇒20s= 188-100
∴ s= 4.4
Hence the value of KN is 4.4
Well, we could try adding up odd numbers, and look to see when we reach 400. But I'm hoping to find an easier way.
First of all ... I'm not sure this will help, but let's stop and notice it anyway ...
An odd number of odd numbers (like 1, 3, 5) add up to an odd number, but
an even number of odd numbers (like 1,3,5,7) add up to an even number.
So if the sum is going to be exactly 400, then there will have to be an even
number of items in the set.
Now, let's put down an even number of odd numbers to work with,and see
what we can notice about them:
1, 3, 5, 7, 9, 11, 13, 15 .
Number of items in the set . . . 8
Sum of all the items in the set . . . 64
Hmmm. That's interesting. 64 happens to be the square of 8 .
Do you think that might be all there is to it ?
Let's check it out:
Even-numbered lists of odd numbers:
1, 3 Items = 2, Sum = 4
1, 3, 5, 7 Items = 4, Sum = 16
1, 3, 5, 7, 9, 11 Items = 6, Sum = 36
1, 3, 5, 7, 9, 11, 13, 15 . . Items = 8, Sum = 64 .
Amazing ! The sum is always the square of the number of items in the set !
For a sum of 400 ... which just happens to be the square of 20,
we just need the <em><u>first 20 consecutive odd numbers</u></em>.
I slogged through it on my calculator, and it's true.
I never knew this before. It seems to be something valuable
to keep in my tool-box (and cherish always).
