OK I will answer give me a few minutes I’ll try to be fast and come back to you in the comment section
Answer:
The even numbers between 0 and X represents an arithmetic sequence with a common difference of 2
The rule of arithmetic sequence = a + d(n - 1)
Where a is the first term and n is the number of terms
So, for the even numbers between 0 and X
The first term = a = 0
d = 2
So, we need to find n at the last term which is X
∴ X = 0 + 2 ( n -1 )
∴ n - 1 = X/2
∴ n = X/2 + 1
The sum of the arithmetic sequence = (n/2) × (2a + (n−1)d)
Substitute with a and d and X
So, the sum = (n/2) * (2*0 + (n−1)*2)
= (n/2) * ((n−1)*2)
= n(n-1)
= (X/2 + 1) * (X/2)
= X/2 by (X/2 + 1)
So, The quick way to add all even numbers between 0 and X always works.
Answer:
5/8
Step-by-step explanation:
Answer:
That sucks.
Step-by-step explanation:
MrBillDoesMath!
Answer to #4: 81/256 * s^8 * t^ 12
Comments:
(7x^3) ^ (1/2) = 7 ^ (1/2) * x^(3/2) where ^(1/2) means the square root of a quantity. The answer written (7x^3) is NOT correct.
---------------------
(1) (27s^7t^11)^ (4/3)
= 27^(4/3) * (s^7)^(4/3) * (t^11)^ (4/3)
As 27 = 3^3, 27 ^(4/3) = 3^4 = 81
(2) (-64st^2)^ (4/3) = (-64)^(4/3) * (s^4/3) * t(^8/3)
As 64 = (-4)^3, (-64)^(4/3) = (-4)^4 = +256
So (1)/(2) =
81 * s^(28/3)* t^(44/3)
------------------------------- =
256 s^(4/3) * t^((8/3)
81/256 * s ^ (28/3 - 4/3) * t^(44/3 - 8/3) =
81/256 * s^(24/3) * t (36/3) =
81/256 * s^8 * t^ 12
MrB
