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1 answer:
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Answer:
CD ≠ EF
Step-by-step explanation:
Using the distance formula
d =
with (x₁, y₁ ) = C(- 2, 5) and (x₂, y₂ ) = D(- 1, 1)
CD =
=
= =
Repeat using (x₁, y₁ ) = E(- 4, - 3) and (x₂, y₂ ) = F(- 1, - 1)
EF =
=
= =
Since ≈
, then CD and EF are not congruent
Answer:
YES
NO
NO
Step-by-step explanation:
The given polynomial is:
(x - a) is a factor of a polynomial iff x = a is a solution to the polynomial.
To check if (x - 5) is a factor of the polynomial f(x), we substitute x = 5 and check if it satisfies the equation.
∴ f(5) = 5³ + 4(5)² - 25(5) - 100
= 125 + 100 - 125 - 100
= 225 - 225
= 0
We see, x = 5 satisfies f(x). So, (x - 5) is a factor to the polynomial.
Now, to check (x + 2) is a factor.
i.e., to check x = - 2 satisfies f(x) or not.
f(-2) = (-2)³ + 4(-2)² - 25(-2) - 100
= -8 + 16 + 50 - 100
= -108 + 66
≠ 0
Therefore, (x + 2) is not a factor of f(x).
To check (x - 4) is a factor.
∴ f(4) = 4³ + 4(4)² - 25(4) - 100
= 64 + 64 - 100 - 100
= 128 - 200
≠ 0
Therefore, (x - 4) is not a factor of f(x).