-77
16t(t + 4) is the factorization Usually it's set = to 0 16t (t+4) =0 So, either 16t = 0 or t + 4 = 0 t = 0 or t = -4
t with a small 4 to the top right of it
Neither, it is a power sequence. t(n) = n4 so t(6) would be 64 = 1296
To calculate ( t ) raised to the power of ( n ) in base ( 4 ) (denoted as ( t^{n4} )), you first convert ( n ) to its base ( 4 ) representation. Then, evaluate ( t ) raised to the exponent that corresponds to this base ( 4 ) representation. Essentially, you would compute ( t^{n_0 \cdot 4^0 + n_1 \cdot 4^1 + n_2 \cdot 4^2 + \ldots} ), where ( n_0, n_1, n_2, \ldots ) are the digits of ( n ) in base ( 4 ). Finally, you can simplify the expression for the final result.
x2 is the same as x times x. In this case x = t+2 so we can say (t+2)2 is (t+2)(t+2) or t2+4t+4
t^4 - 81 = (t^2)^2 - (3^2)^2 = (t^2 - 3^2)(t^2 + 3^2) = (t - 3)(t + 3)(t^2 + 9)
81.
16t(t + 4) is the factorization Usually it's set = to 0 16t (t+4) =0 So, either 16t = 0 or t + 4 = 0 t = 0 or t = -4
T to the fourth power is already fully simplified unless you have a value for "t."
You can take a 5s2t out of all of that, leaving yourself with 3s4t3 + 19s3t2 + 7s2t4 - 7
t with a small 4 to the top right of it
1
The prime factorization for 68 is: 2 x 2 x 17
(t - 3)(t + 3)(t2 + 9)
Irving Layton died on January 4, 2006 at the age of 93.
t4-81 is a difference of 2 squares and can be written as (t2-9)(t2+9) t2+9 can't be further factorised but t2-9 is a difference of 2 squares again and can be factorised to (t+3)(t-3) so the factors of t4-81 are :(t2+9)(t+3)(t-3) Hope this helps :-) I believe the answer you are looking for is (t - 3)(t + 3)(t 2 + 9)
Neither, it is a power sequence. t(n) = n4 so t(6) would be 64 = 1296