There may be more brilliant solution than the following, but here are my thoughts.
We make use of Euclid's algorithm to help us out.
Consider finding the hcf of A=2^(n+x)-1 and B=2^(n)-1.
If we repeated subtract B from A until the difference C is less than B (smaller number), the hcf between A and B is the same as the hcf between B and C.
For example, we would subtract 2^x times B from A, or
C=A-2^xB=2^(n+x)-2^x(2^n-1)=2^(n+x)-2^(n+x)+2^n-1=2^n-1
By the Euclidean algorithm,
hcf(A,B)=hcf(B,C)=hcf(2^n-1,2^x-1)
If n is a multiple of x, then by repetition, we will end up with
hcf(A,B)=hcf(2^x-1,2^x-1)=2^x-1
For the given example, n=100, x=20, so
HCF(2^120-1, 2^100-1)=2^(120-100)-1=2^20-1=1048575
(since n=6x, a multiple of x).
1A.) Cannot Be Factored
2B.) 7n(2n + 7 + n)
3C.) (b+3) (2b + 5)
4D.) -(xy - 2x + 12)
5E.) (r-5)(7r^2+6)
Answer:
EF = 6
Step-by-step explanation:
Using the secant- secant power theorem, that is
GF × GE = GH × GS
4(4 + 2x) = 5(5 + x) ← distribute both sides
16 + 8x = 25 + 5x ( subtract 5x from both sides )
16 + 3x = 25 ( subtract 16 from both sides )
3x = 9 ( divide both sides by 3 )
x = 3
Then
EF = 2x = 2 × 3 = 6
Answer:
YES
Step-by-step explanation:
To determine a function on a graph use the vertical line test!
put a vertical line through the graph at ANY POINT and if it only passes through one point then it is a function.
