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andrew-mc [135]
3 years ago
7

State how many imaginary and real zeros the function has.

Mathematics
1 answer:
myrzilka [38]3 years ago
7 0
Hello,
Degre 5==> even number of imaginary roots and odd number of reals.

Real roots must be a divisor of 7.
Using Excel, we find


<span> <span><span> x  P(x) 
 </span><span>-7 0 </span>
<span> -6 1369
</span> <span> -5 1352
</span> <span> -4 867
</span> <span> -3 400
</span> <span> -2 125
</span> <span> -1 24
 </span><span>0 7
 </span><span>1 32
 </span><span>2 225
 </span><span>3 1000
 </span><span>4 3179
 </span><span>5 8112
 </span><span>6 17797
 </span><span>7 35000

So: one real root: 7 and 4 imaginary roots.

Answer B
 </span></span></span>
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Please help!! And pls show work
coldgirl [10]
<h3>The answer to your question is k (-3) = 21!</h3>

Here's how I got this answer:

<em><u>K(a) = 2a^2 - a </u></em>

<em><u>K(a) = 2a^2 - a K(-3) = 2(-3)^2 - (-3)</u></em>

<em><u>K(a) = 2a^2 - a K(-3) = 2(-3)^2 - (-3)K(-3) = 2(9) + 3</u></em>

<em><u>K(a) = 2a^2 - a K(-3) = 2(-3)^2 - (-3)K(-3) = 2(9) + 3K (-3) = 18 + 3</u></em>

<em><u>K(a) = 2a^2 - a K(-3) = 2(-3)^2 - (-3)K(-3) = 2(9) + 3K (-3) = 18 + 3K (-3) = 21</u></em>

I hope this helps!

Also sorry for the late answer, I just got the notification that you replied to my comment, I hope I came in time!

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Answer:

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