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What are the factors of t4 - 81?
t^4 - 81 = (t^2)^2 - (3^2)^2 = (t^2 - 3^2)(t^2 + 3^2) = (t - 3)(t + 3)(t^2 + 9)
H equals 16t2 plus 64t?
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
How do you simplify t 4 power?
T to the fourth power is already fully simplified unless you have a value for "t."
What is t to the fourth power?
t with a small 4 to the top right of it
How do you simplify t-4 power?
1
What is 15sto the 6 power t to the 4 power plus 95s to 5 power t to the 3 power plus 35 s to the 4 power t to the 5 power minus 35 s to the 2 power t divided by 5 s to the 2 power t?
You can take a 5s2t out of all of that, leaving yourself with 3s4t3 + 19s3t2 + 7s2t4 - 7
T is the prime factorization for 68?
The prime factorization for 68 is: 2 x 2 x 17
What is the factorization of 2t plus 20?
2(t + 10)
What is the factors of t4-81?
(t - 3)(t + 3)(t2 + 9)
What are the factors of t4-81?
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)
What is t sum of 2 to the power of 4 and 7 to the power of 2 Write the answer in standard form?
16 + 49 = 65
When did Edwin T. Layton die?
Edwin T. Layton died on April 12, 1984 at the age of 81.
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