This is the quickest way: 1. The number must be ending with 5 as original number is ending with 25 2. by inspection it is a 5 digit number begining with 3 and ending with 5. That is 3xyz5. 3. The number barred the 25 (ie 11112222) must be a product of two consecutive numbers. 4. 11112222=1111x10002 (easy to see from pattern) = 1111x3334x3=3333x3334 5. The number = 33335 (rule of square of a number ending with 5)
Menurut saya, untuk akar kwadrat itu lebih mudah dan lebih sederhana kalau DIHITUNG LANGSUNG saja dengan menerapkan bahwa (a + b)² = a² + 2ab + b² = a² + b(2a +b). Dengan mengambil setiap tahap 2 digit dihitung dari depan kebelakang. Dengan cara itu bahkan bisa untuk menghitung angka pecahan desimal.
Actually there's a pattern with n number of 3s and 5 Like 35² is One 1 two 2 1225 335²= Since there are two threes we have two ones and three twos 112225 And here we have four ones So four threes 33335 Done! Like if useful
I enjoyed watching this. Calm and infirmatuve. 😊
This is the quickest way:
1. The number must be ending with 5 as original number is ending with 25
2. by inspection it is a 5 digit number begining with 3 and ending with 5. That is 3xyz5.
3. The number barred the 25 (ie 11112222) must be a product of two consecutive numbers.
4. 11112222=1111x10002 (easy to see from pattern) = 1111x3334x3=3333x3334
5. The number = 33335 (rule of square of a number ending with 5)
by inspection ,the square near 11 is 9 ,the last digit ends at 5 must be ,so the answer is 33335. Easy.
Menurut saya, untuk akar kwadrat itu lebih mudah dan lebih sederhana kalau DIHITUNG LANGSUNG saja dengan menerapkan bahwa (a + b)² = a² + 2ab + b² = a² + b(2a +b). Dengan mengambil setiap tahap 2 digit dihitung dari depan kebelakang. Dengan cara itu bahkan bisa untuk menghitung angka pecahan desimal.
Did you know from the beginning that it could be applied to the formula for squaring (a+b)?
Continue sem desistir.
Actually there's a pattern with n number of 3s and 5
Like 35² is
One 1 two 2
1225
335²=
Since there are two threes we have two ones and three twos
112225
And here we have four ones
So four threes 33335
Done!
Like if useful
❤❤🎉
1111222225 = 1111222200 + 25
= 11110 * 100020 + 25
= 11110 * （99990 + 30） + 25
= a * (9a + 30) + 25
= 9a^2 + 30a + 25
= (3a + 5)^2
= 33335^2
I did that shit in my head, dunce.
let m=11111, 10^5=99999+1=9m+1
1111222225=(m+1)*(9m+1)+2m+3=9m^2+12m+4=(3m+2)^2
(1111222225)^(1/2)=3m+2=33333+2=33335
33335
Not fit for compitative exams
my calculator can do this in 3 seconds
@@jetstream-h6p ever heard of efficency, kinda what evolution banks on
@@jetstream-h6p so how about you keep your words to yourself and grow up
This is a maths problem, not an arithmetic problem.
@@brodieanderson9598 Why don't you use chatGPT to comment on this page. It is not a sentence,but just an enume ration of words.
You should get a faster calculator.