how do you solve 6 radical 343 subtract 2 radical 7?

Answer 1

6*sqrt(343) = 6*sqrt(49)*sqrt(7) = 6*7sqrt(7)

42sqrt(7) - 2sqrt(7) = 40sqrt(7)

Answer 2

First, I would try to factor the 343. Since the second radical is radical 7, try to divide 343 by 7

343 = 7*49
So you have:
6 radical (7*49) - 2 radical 7=
6(7) radical 7 - 2 radical 7 =
42 radical 7 - 2 radical 7=
40 radical 7

Answer 3

6 radical (343) - 2 radical (7) =
6 radical (49*7) - 2 radical (7) =
6*7 radical (7) - 2 radical (7) =
42 radical (7) - 2 radical (7) =
40 radical (7)

Answer 4

6[ radical 49 * radical 7]- 2 radical 7=
42 radical 7- 2 radical 7= 40radical 7

Answer 5

6√(343) - 2√(7)
Since 343 = 7³, the problem becomes easier.
6√(343) - 2√(7)
6√(49*7) - 2√(7)
6√(49)√(7) - 2√(7)
6(7)√(7) - 2√(7)
42√(7) - 2√(7)
40√(7)

Answer 6

To solve this, you want to take the factors of 343 (in pairs) and determine which has the highest perfect square number. Since you have the terms 6 sq. rt. 343 and 2 sq. rt. 7, you might want to divide 343 by 7. By doing so, you get 49, which is indeed the highest perfect square of 343. The square root of 49 is seven. You multiply the seven by 6 to get a new term of 42 sq. rt. 7. Then you add this to 2 sq. rt. 7. Your final answer is 44 sq. rt. 7.
Hope this helps!

Answer 7

a million) ?(-a hundred twenty 5/343) = ?(-5^3/7^3) = (-5^(3*a million/3)/7^(3*a million/3) = -5/7 2) ?(36g^6) = (36g^6)^a million/2 = 6^(2*a million/2)g^(6*a million/2) = 6g^3 3) ?((128^thirteen)(b^6)) = (128^thirteen)b^6)^(a million/3) = 128^(thirteen/3)b^2 want this enables.

Answer 8

6 sqrt()343-2 sqrt(7)=6*7sqrt(7)-2sqrt(7)=40*sqrt)(7).=105.83 answer

Answer 9

6root343 - 2root7 =

6root[(7)(49)] - 2root7 =

6[(root7)(root49)] - 2root7 =

(root7)[(6root49) - 2] =

(root7)[6(7) - 2] =

(root7)(42 - 2) =

40root7.