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Andre45 [30]
3 years ago
14

A fourth degree polynomial with real coefficients can have -4, 8i,+4i, and as its zeros. True or false? Justify your answer.

Mathematics
1 answer:
liraira [26]3 years ago
4 0
You didn't give the fourth zero, but the answer is still false. If you have a root or an imaginary number as a zero, then its conjugate is also a zero. So if 8i is a zero, then -8i must also be a zero, and if 4i is a zero, then -4i must be a zero, with those zeros and -4, the number of zeroes exceeds the number of zeroes that a fourth degree polynomial can have.
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Find the point P on the line yequals=66x that is closest to the point (74 comma 0 )(74,0). What is the least distance between P
tekilochka [14]

Answer:

The point P that is closest to the point (74, 0) is P(7.4, 22.2), and the least distance between P and (74, 0) is 70.2

Step-by-step explanation:

Given

y = 66x

And a point A(74, 0)

We are required to find the point on the line that is closest to A, and then find the distance between these points.

Let the point be P(s, t)

Because it's coordinate satisfies the given equation

y = 3x

Let t = 3s, then we have

p(s, t) = P(s, 3s)

Next, we find the distance between P and A

Let the distance be

D = √[(x1 - y1)² + (x2 - y2)²]

Here x1 = 74, x2 = 0, y1 = s, y2 = 3s.

D = √[(74 - s)² + (0 - 3s)²]

= √(s² - 148s + 5476 + 9s²)

= √(10s² - 148s + 5479)

To maximize the distance, we find the derivative of D with respect to s, and set it to 0.

That is dD/ds = 0

dD/ds =

(20s - 148)/2√(10s² - 148s 5479) = 0

=> 20s - 148 = 0

20s = 148

s = 148/20

s = 7.4

Since t = 3s, we have

t = 22.2

So, the point is P(7.4, 22.2)

Finally, we find the distance D.

D = √[(74 - s)² + (0 - 3s)²]

D = √[(74 - 7.4)² + (0 - 3×7.4)²]

D = √(4435.56 + 492.84)

= √4928.4

D = 70.2

Therefore, the distance is 70.2

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