Find the image of O(–2, –1) after two reflections, first across the line y = –5, and then across the line x = 1.
2 answers:
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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.
Answer:
Step-by-step explanation:
P = x^2 + 6
Solving for 4x^2:
P = x^2 +6
P - 6 = x^2
4(P - 6) = 4x^2
Substitution: