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avanturin [10]
3 years ago
6

Drag the expressions into the boxes to correctly complete the table.

Mathematics
1 answer:
Maslowich3 years ago
3 0

Answer:

Polynomial

x^4 +3x^3 -5x^2 -7x +20\\\\x^3 -2x^2 +x -12\\\\x^2 + 3x^5 -6x +12x^3

not a polynomial

4\sqrt[3]{x} -\sqrt{x} -20\\\\x^{-3}-x}^{-2}-6x^{-1}+8\\\\\frac{5}{x^3} -\frac{1}{x^2} +\frac{8}{x} -40\\\\

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Step-by-step explanation:

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3 0
3 years ago
TanA+ secA -1/ tanA - secA+1= 1+ sinA/ cosA
pantera1 [17]

Answer:

See below.

Step-by-step explanation:

I can do this but it's a pretty long proof. There might be a much easier way of proving this but this is the only way I can think of.

Write tan A as s/c and sec A as 1/c ( where  s and c are  sin A  and cos A respectively).

Then tanA+ secA -1/ tanA - secA+1

= (s/c + 1/c - 1) / ( sc - 1/c + 1)

= (s + 1 - c) / c  /   (s - 1 + c) / c

= (s - c + 1) / (s + c - 1).

Now we write the right side of the identity  ( 1 + sin A) / cos A as  (1 + s) / c

So if the identity is true then:

(s - c + 1) / (s + c - 1) = (1 + s) / c.

Cross multiplying:

cs - c^2 + c =  s + c - 1 + s^2 + cs - s

Simplifying:

cs - c^2 + c = cs - (1 - s^2)  + c + s - s

Now the s will disappear on the right side and 1 - s^2 = c^2 so we have

cs - c^2 + c = cs - c^2 + c.

Which completes the proof.

8 0
3 years ago
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