The correct rectangular equivalence of 3sqrt(2)·cis(7pi/4 ) is:
3sqrt(2)·cos( 7pi/4 ) + i·sqrt(2)·sin( 7pi/4 ) = 3 - 3i.
<h3>Where did David go wrong?</h3>
David mistakenly interchanged the Sin function and the Cos function when he was calculating the problem.
Hence the correct rectangular equivalence is:
3sqrt(2)·cos( 7pi/4 ) + i·sqrt(2)·sin( 7pi/4 ) = 3 - 3i.
<h3>What is rectangular equivalence?</h3>
An equation is rectangular in form when it is comprised of Variables like X and Y and can be represented on a Cartesian Plane.
Learn more about rectangular equivalence at:
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If you're asking for extrema, like the previous posting
well

like the previous posting, since this rational is identical, just that the denominator is negative, the denominator yields no critical points
and the numerator, yields no critical points either, so the only check you can do is for the endpoints, of 0 and 4
f(0) = 0 <---- only maximum, and thus absolute maximum
f(4) ≈ - 0.19 <---- only minimum, and thus absolute minimum
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).
Its right.................
Answer:
No solutions. You can’t have a negative absolute value.