61 squared equals 3,721.
It must be 61 square inches since that bit of information cannot relate to anything else.
The squared area of the box, from the question itself, is 61 square inches. Scaling up the linear dimensions by a factor of 10 will make the area 6100 square inches.
58 * * * * * No. Distance squared = (-2 - 3)2 + (1 - 7)2 = 52 + 62 = 25 + 36 = 61 So distance = sqrt(61) = 7.8 approx.
9x squared-15x squared = -6
1 squared = 1 2 squared = 4 3 squared = 9 4 squared = 16 5 squared = 25 6 squared = 36 7 squared = 49 8 squared = 64 9 squared = 81 10 squared = 100
It must be 61 square inches since that bit of information cannot relate to anything else.
The squared area of the box, from the question itself, is 61 square inches. Scaling up the linear dimensions by a factor of 10 will make the area 6100 square inches.
There are no two whole numbers which when squared sum to 62. There are infinitely may pairs of irrational numbers that when squared sum to 62, eg 1 and √61, 2 and √58
58 * * * * * No. Distance squared = (-2 - 3)2 + (1 - 7)2 = 52 + 62 = 25 + 36 = 61 So distance = sqrt(61) = 7.8 approx.
The pattern is the squared numbers with the digits reversed. 92 = 81 → 18 82 = 64 → 46 72 = 49 → 94 62 = 36 → 63 52 = 25 → 52 42 = 16 → 61
he made the theorem C squared = A squared + B squared and A squared = C squared - B squared or B squared = C squared - A squared
If yoiu mean pt A = ( 8,2) & B = (3,8) We apply Pythagoras. Distance 'd^(2)' = (x(A) - x(B))^2 + (y)A) - y(B))^(2) Substituting d^)2 = (8 - 3)^(2) + ( 2 - 8)^(2) d^(2) = 5^(2) + -6^(2) d^(2) = 25 + 36 d^(2) = 61 d = sqrt(61)
between 5 and 6 (5 squared is 25 and 6 squared is 36)
9x squared-15x squared = -6
2 squared X 3 squared = 5 squared
1 squared = 1 2 squared = 4 3 squared = 9 4 squared = 16 5 squared = 25 6 squared = 36 7 squared = 49 8 squared = 64 9 squared = 81 10 squared = 100
100cm squared = 0.01 metres squared