Let f(x)=xn +a1xn−1 +a2xn−2 +...+an,
where a1, a2, . . . , an are given numbers. It is given that f(x) can be written in the form
f (x) = (x + k1 )(x + k2 ) · · · (x + kn ). By considering f(0), or otherwise, show that k1k2 ...kn = an.
Show also that
(k1 +1)(k2 +1)···(kn +1)=1+a1 +a2 +...+an and give a corresponding result for (k1 − 1)(k2 − 1) · · · (kn − 1). (5)
Hint: For the last part, consider the cases n odd and n even separately.

hence, the required result is

step-by-step explanation:

suppose we have two matrices a and b

we will assume the two matrices

we have to compute b-a which means subtracting corresponding elements of a from b.

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answer:   the paragraph proof is mentioned below.

step-by-step explanation:

given: in a δxwy where

such that, ∠wzx ≅ ∠yzx and zw≅zy

we have to prove that: segment zx is a perpendicular bisector of segment wy.

prove: since, zw≅zy ⇒ ∠wzy is a supplementary angle.

then, ∠wzy= 180°

⇒∠wzx + ∠yzx = 180° ( because xz divides ∠wzy into two parts)

but, ∠wzx ≅ ∠yzx ( given)

⇒∠wzy+∠wzy=180° ( by substitution)

⇒2∠wzy=180°

⇒∠wzy=90°

⇒∠wzy=∠yzx =90°

therefore, xz⊥wy where wz=zy

⇒zx is a perpendicular bisector of wy.

answer:

probably the same as it is today since you did not eat the corn dogs yet

step-by-step explanation: