226
first you multiply 33333333334 by 33333333334 (if its kinda hard to count these, they both have ten 3's, and one 4) and then you should get this: 1111111111155555555556 (for this one, there are eleven 1's, ten 5's, and one 6) then you add all these digits together, and should get this: 67 ***p.s. the reason why it's not 1156 is because, in the order of operations, you have to solve the exponents (to the second power) before you can start adding***
go to http://www.examscouncil.or.ke or on your phone type KCSE followed by the nine digits of the candidate's index number e.g. KCSE 713403xxxthen send to 2228 for both zain and safaricom charged at 20 bob per message
go to http://www.examscouncil.or.ke or on your phone type KCSE followed by the nine digits of the candidate's index number e.g. KCSE 123456789 then send to 2228 for both zain and safaricom charged at 20 bob per message
Straight line depreciation method allocate equal amount for all years while in sum of years digit method depreciation is allocated with high amount in initial years while low amount in later years.
I've been trying to figure this out myself for a discrete structure class.From the digging I've done online I've found this formula thus far:if n is the number of digits, and s is the sum:C(n+s-1 , n-1) where C denotes "choose" as in C(n , k) "n choose k" which can be solved by(n! / [ (n-k)! * k! ] )this seems to work for situations where the sum is < 10, or so claims the forum I found.I whipped it all up into a scheme function if anyone wants to take advantage:;;factorial: num -> num;;finds factorial of num.(define (factorial num)(if (or (= num 1) (= num 0))1(* num (factorial (sub1 num)))));;==========================;;xchy: num num -> num;;the "n choose k" function(define (xchy n k)(/ (factorial n)(* (factorial(- n k))(factorial k))));;==========================;;n-dig&sum-s: num num -> num;;finds the number of n digit combinations with sum s;;C(n+s-1,n-1)(define (n-dig&sum-s n s)(xchy (sub1 (+ n s))(sub1 n)))
First, 54 rounded to the nearest hundred is: 100.We round the number up to the nearest hundred if the last two digits in the number are 50 or above.
5600
The number would be 2666. 6 is repeated 3 times. Rounded to the nearest hundred is 2700 and rounded to the nearest thousand is 3000.
199
27222722
278879 rounded to the nearest hundred - is 278900. The last two digits (79) are greater than 50 so you round up.
863.00
2722
To round to the nearest hundred and ten both being 200 it leaves 9 possibilities. 195,196,197,198,199,201,202,203, and 204. The only one with digits adding up to 19 is 199.
listen tell me the thing is is that we not really paying attention like that listen if you can't get your answer up to 200 you will know that if you add your tan to something that's in his hundred like if you do that I'll get you to your answer this if I had say if I had 198 we post it rounded to nearest ten
Remember, we did not necessarily round up or down, but to the hundred that is nearest to 321. First, 321 rounded to the nearest hundred is: 300 When rounding to the nearest hundred, like we did with 321 above, we use the following rules: A) We round the number up to the nearest hundred if the last two digits in the number are 50 or above. B) We round the number down to the nearest hundred if the last two digits in the number are 49 or below. C) If the last two digits are 00, then we do not have to do any rounding, because it is already to the hundred.
The number 47 does not have a digit in the hundreds place. It only has digits in the ones and tens places. The four is in the tens place and the seven is in the ones place. Therefore it cannot be rounded to the nearest hundreds place. The number 47 is between zero hundred and one hundred. It is nearer to zero hundred than it is to one hundred. When rounded to the nearest hundred, it becomes zero, since zero hundred is nearer to it than one hundred is.