If s and t are integers greater than 1 and each is a factor of the integer n, which of the following must be a factor of n^{st}
?
1) s^t
2) (st)²
3) s+t
A) None
B) 1 only
C) 2 only
D) 3 only
E) 1 and 2
2 answers:
Answer:
E ( 1 and 2) only
Step-by-step explanation:
s is an integer and a factor of n and is greater than 1 and t is also an integer and a factor of and equally greater than 1
s^t is a factor n^(st)
let s = 3 and t = 4 and n = 12
then n^(3×4) = 12 ^12= (3 × 4) ^(3×4) = 3³ × 4⁴ or 3⁴ × 4³
(st)² = (3×4)² = (12)² which is a factor of 12¹²
3^4
s + t = 3+4 = 7 and 7 is not a factor 12¹²
let say s = 2, and t = 3
2 + 3 = 5 and 5 is not a factor of 6⁶
Answer:
E) 1 and 2
Step-by-step explanation:
We are given that there are two integers (s and t) and they are factors of another integer (n). For example if s = 3 and t = 2, we can have n = 6.
Thus:
n^(st) = 6^(2*3) = 6^6 = (2^6)(3^6)
For the first condition: s^t = 3^2 is a factor of (2^6)(3^6)
For the second condition: (st)^2 = (3*2)^2 = 6^2 is a factor of 6^6
For the third condition: s+t = 3+2 = 5 is not a factor of 6^6 or (2^6)(3^6)
Therefore, only 1 and 2 are factors of n^(st)
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Answer:
four million
Step-by-step explanation:
just subtract 12 million by 8 million
4/10 = 40% of throws made
40% of 24 = 9.6 which rounds up to 10
The answer got to be D. 1/5. Because 1/5=the same thing as 5.
Hope that helps!!
<h3>
Answer: 28k+68</h3>
=============================================
Work Shown:
7(8+4k)+12
7*8+7*4k+12 ..... distributive property
56+28k+12
28k+56+12
28k+68
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