Find the H. C.F for 513, 1134 and 1215

Solution:

In this question no method is specified to find the H.C.F so

Here we find the H.C.F by Euclid division lemma:-

a=bq+r

Here a= 1215 , b = 1134

1215>1134>513

1215=1134(1)+81

1134=81(14)+0

remainder becomes 0,the Hcf of 1134 and 1215 is 81.

Then find the Hcf of 81 and 513

513=81(6)+27

81=27(3)+0

remainder becomes 0, HCF of 513 & 81 is 27.

Hence, the HCF of 513,1134 and 1215 is 27.

By Prime factorization method

1215= 3×3×3×3×3×5= 3^5 ×5¹

1134= 3×3×3×3×2×7= 3⁴×2¹×7¹

513= 3×3×3×19= 3³× 19¹

H.C.F= 3³= 3×3×3= 27

Hence, the HCF of 513,1134 and 1215 is 27.

Hope this will help you

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2

describing distributions is one of the key skills you’ll need to earn a high score on the ap statistics exam. if you need proof of this, just flip through some past exam questions, which can be found at the collegeboard website. you’ll notice that the first free response question is almost always a question requiring you to look at a graph and describe it.

step-by-step explanation: just trying to get points sorry : (