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lesya [120]
3 years ago
8

Polynomial Functions help

Mathematics
1 answer:
Studentka2010 [4]3 years ago
6 0
The key piece of information for these questions is the Fundamental Theorem of Algebra, which states that a degree n polynomial has n complex roots. A complex root can be either real or imaginary. 

First question, regarding the polynomial y = x^3 - 3x^2 + 16x - 48:
We know there is one real root, the x-intercept.
Since it's a third degree polynomial, there are three complex roots in total. 
Therefore, there is one real root and two imaginary roots.
Answer is B

Second question:
You probably can guess the answer, now that you know the Fundamental Theorem of Alegebra:
There are 3 real zeros, each with multiplicity one, meaning each root only happens once. It's a 5th degree polynomial, so there are a total of 5 roots, implying 2 imaginary roots. 

Answer is C) 3 real and 2 imaginary zeroes.

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Katyanochek1 [597]

Write tan in terms of sin and cos.

\displaystyle \lim_{t\to0}\frac{\tan(6t)}{\sin(2t)} = \lim_{t\to0}\frac{\sin(6t)}{\sin(2t)\cos(6t)}

Recall that

\displaystyle \lim_{x\to0}\frac{\sin(x)}x = 1

Rewrite and expand the given limand as the product

\displaystyle \lim_{t\to0}\frac{\sin(6t)}{\sin(2t)\cos(6t)} = \lim_{t\to0} \frac{\sin(6t)}{6t} \times \frac{2t}{\sin(2t)} \times \frac{6t}{2t\cos(6t)} \\\\ = \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right)

Then using the known limit above, it follows that

\displaystyle \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right) = 1 \times 1 \times \frac3{\cos(0)} = \boxed{3}

4 0
1 year ago
1 ) Considere a expressão algébrica:
balandron [24]
A) i think is correct andewmer
7 0
3 years ago
HELP IM DUM lol
Ghella [55]

Answer:

89%

Step-by-step explanation:

89 out of one hundred is shaded orange to it would be 89/100 and 89/100 is equal to 89%.

6 0
3 years ago
Read 2 more answers
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