A x = 0
using the law of exponents = 1
for (6² )^ x = 1 then x = 0
B note that = 1 ⇒ x = 1
2 → 2^8 × 3^(-5) × 1^(-2) × 3^(-8) × 2^(-12) × 2^(28)
= 2^(8 -12 + 28) × 1 × 3^(- 5 - 8)
= 2^24 × 3^(- 13) = 2^(24)/3^(13) = 10.523 ( 3 dec. places)
I've included two questions. Hopefully you can help me with both. Thanks!
1 answer:
Answer:
1. value is 0; x-3 is a factor . . . . . . . . . . . . . .third choice
2. evaluates at x = -1; remainder is -11 . . . . first choice
Step-by-step explanation:
Dividing f(x) by (x -a) gives ...
f(x)/(x -a) = g(x) +r/(x -a) . . . . some quotient and a remainder r
If we multiply this expression by (x -a), we see ...
f(x) = (x -a)g(x) +r
so
f(a) = (a -a)g(a) +r . . . . . evaluate the above equation at x=a
f(a) = 0 +r
f(a) = r . . . . . . . . . a statement of the remainder theorem
If r=0, then x-a is a factor of f(x) = (x-a)g(x).
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1. We have "a" = 3, and f(3) = 0. Therefore (x-3) is a factor.
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2. We have "a" = -1, and f(-1) = -11. Therefore the remainder from division by (x+1) is -11.
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