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Sidana [21]
3 years ago
10

How many complex roots does the polynomial equation below have?

Mathematics
1 answer:
erastova [34]3 years ago
5 0

Answer:

4.

Step-by-step explanation:

I will have one real root = fifth root of 3.

As it has a total of 5 roots ( By the Fundamental Theorem of Algebra) then it must have 4 complex roots.

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The ratio of boys to girls in Mr. Hakeem's class is 64. If there are 18 boys, how many girls
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Answer: C. 12 girls

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6 0
3 years ago
What is the radius and diameter of 7
Aleks04 [339]
<h3>Answer:</h3>

The radius would be <em>3.5 units</em>.

Step-by-step explanation:

This is because a radius is always half of a diameter.

By using that,

7 ÷ 2 = 3.5 units

Since there was no measurement listed in the question, the answer would be in "units."

8 0
3 years ago
The plane x+y+2z=8 intersects the paraboloid z=x2+y2 in an ellipse. Find the points on this ellipse that are nearest to and fart
DiKsa [7]

Answer:

The minimum distance of   √((195-19√33)/8)  occurs at  ((-1+√33)/4; (-1+√33)/4; (17-√33)/4)  and the maximum distance of  √((195+19√33)/8)  occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

Step-by-step explanation:

Here, the two constraints are

g (x, y, z) = x + y + 2z − 8  

and  

h (x, y, z) = x ² + y² − z.

Any critical  point that we find during the Lagrange multiplier process will satisfy both of these constraints, so we  actually don’t need to find an explicit equation for the ellipse that is their intersection.

Suppose that (x, y, z) is any point that satisfies both of the constraints (and hence is on the ellipse.)

Then the distance from (x, y, z) to the origin is given by

√((x − 0)² + (y − 0)² + (z − 0)² ).

This expression (and its partial derivatives) would be cumbersome to work with, so we will find the the extrema  of the square of the distance. Thus, our objective function is

f(x, y, z) = x ² + y ² + z ²

and

∇f = (2x, 2y, 2z )

λ∇g = (λ, λ, 2λ)

µ∇h = (2µx, 2µy, −µ)

Thus the system we need to solve for (x, y, z) is

                           2x = λ + 2µx                         (1)

                           2y = λ + 2µy                       (2)

                           2z = 2λ − µ                          (3)

                           x + y + 2z = 8                      (4)

                           x ² + y ² − z = 0                     (5)

Subtracting (2) from (1) and factoring gives

                     2 (x − y) = 2µ (x − y)

so µ = 1  whenever x ≠ y. Substituting µ = 1 into (1) gives us λ = 0 and substituting µ = 1 and λ = 0  into (3) gives us  2z = −1  and thus z = − 1 /2 . Subtituting z = − 1 /2  into (4) and (5) gives us

                            x + y − 9 = 0

                         x ² + y ² +  1 /2  = 0

however, x ² + y ² +  1 /2  = 0  has no solution. Thus we must have x = y.

Since we now know x = y, (4) and (5) become

2x + 2z = 8

2x  ² − z = 0

so

z = 4 − x

z = 2x²

Combining these together gives us  2x²  = 4 − x , so

2x²  + x − 4 = 0 which has solutions

x =  (-1+√33)/4

and

x = -(1+√33)/4.

Further substitution yeilds the critical points  

((-1+√33)/4; (-1+√33)/4; (17-√33)/4)   and

(-(1+√33)/4; - (1+√33)/4; (17+√33)/4).

Substituting these into our  objective function gives us

f((-1+√33)/4; (-1+√33)/4; (17-√33)/4) = (195-19√33)/8

f(-(1+√33)/4; - (1+√33)/4; (17+√33)/4) = (195+19√33)/8

Thus minimum distance of   √((195-19√33)/8)  occurs at  ((-1+√33)/4; (-1+√33)/4; (17-√33)/4)  and the maximum distance of  √((195+19√33)/8)  occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

4 0
3 years ago
2. A man drove 240 miles south at the rate of 60 miles per hour. On the return trip, he drove the rate of 40 miles per hour. The
shutvik [7]

av =  \frac{v_{2} - v_{1}}{t_{2} - t_{1}}

t_{1} =  \frac{d}{v_{1}} =  \frac{240}{60} = 4 \: hours  \\ t_{2} =  \frac{d}{v_{2}}  =  \frac{240}{40}  = 6 \: hours

av =  \frac{v_{2} - v_{1}}{t_{2} - t_{1}}  =  \frac{40 - 60}{6 - 4}   =  \frac{ - 20}{2}  =  - 10

av  \: speed= 10 \\ av \: velocity =  - 10 \:  \: north

3 0
2 years ago
Read 2 more answers
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