Answer:
Step-by-step explanation:
74939e+1364
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.
__
2. We have "a" = -1, and f(-1) = -11. Therefore the remainder from division by (x+1) is -11.
Let G be some point on the diagonal line away from point E.
Angle DEG represents angle 1.
We're given that angle DEF is a right angle which means it's 90 degrees. Angle DEG is some angle smaller than 90 degrees. By definition, that must mean angle 1 is acute. Any acute angle is smaller than 90 degrees. There's not much else to say other than this is just a definition problem.
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Extra side notes:
If angle 1 was a right angle, then that would mean angle GEF would have to be 0 degrees; however the diagram shows this isn't the case.
If angle 1 was obtuse, then there's no way we'd be able to fit it into angle DEF. In other words, there's no way to have an angle larger than 90 fit in a 90 degree angle.
Answer:
Proof in explanation.
Step-by-step explanation:
I'm going to attempt this by squeeze theorem.
We know that
is a variable number between -1 and 1 (inclusive).
This means that
.
for all value
. So if we multiply all sides of our inequality by this, it will not effect the direction of the inequalities.

By squeeze theorem, if 
and
, then we can also conclude that
.
So we can actually evaluate the "if" limits pretty easily since both are continuous and exist at
.

.
We can finally conclude that
by squeeze theorem.
Some people call this sandwich theorem.
Answer:
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