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anzhelika [568]
3 years ago
14

Write the simplest polynomial function for each set of zeros Zeros = -2, 5, -1

Mathematics
1 answer:
deff fn [24]3 years ago
5 0

Answer:

The polynomial is ;

x^3 -2x^2 -13x - 10

Step-by-step explanation:

Given the zeros, we set them to equal zero so as to get the linear factors

Mathematically, that will be as follows;

x = -2

x + 2

x = 5

x-5

x = -1

x + 1

So we have the polynomial as;

(x + 1)(x + 2)(x-5)

= (x -5)(x^2 + 3x + 2)

= x^3+ 3x^2 + 2x -5x^2 -15x - 10

= x^3 -2x^2 -13x - 10

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Kaylis [27]
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7 0
3 years ago
find a set of parametric equations of the line the line pases through the point (2,3,4) and is parallel to the xz-plane and the
joja [24]

Answer:

x=2, y=3, z=4+t

Step-by-step explanation:

For this case we need a line parallel to the plane x z and yz. And by definition of parallel we see that the intersection between the xz and yz plane is the z axis. And we can take the following unitary vector to construct the parametric equations:

u= (u_x, u_y, u_z)= (0,0,1)

Or any factor of u but for simplicity let's take the unitary vector.

Then the parametric equations are given by:

x= P_x + u_x t

y= P_y + u_y t

z= P_z + u_z t

Where the point given P=(2,3,4)= (P_x , P_y, P_z)

And then since we have everything we can replace like this:

x= P_x + u_x t 2+ 0*t = 2

y= P_y + u_y t= 3+ 0*t = 3

z= P_z + u_z t = 4+ 1t = 4+t

x=2, y=3, z=4+t

6 0
3 years ago
How to factor 9X^3-27
azamat
You will find teh GCF of that this will give you 9(x3-3) now you can fator it teh factors of these can be (x2+1) and (x-3)
4 0
3 years ago
there is a traffic light at the intersection of Pine Street and Spruce Avenue. The traffic light on Pine Street follows a cycle.
Alex17521 [72]
The propability is the least chance because it is the lowest time frame that it is on for


8 0
3 years ago
This exercise involves the formula for the area of a circular sector. The area of a circle is 700 m2. Find the area of a sector
mezya [45]

Answer:

334.4 m²

Step-by-step explanation:

The formula for the area of a sector is given as:

1/2 × r² × θ

Where θ = Central angle

Area of a Circle = 700 m²

The formula for the area of a circle = πr²

r = Radius of a circle

r² = Area / π

r = √Area / π

r = √700/π

r = 14.927053304 m

Approximately, r = 14.93 m

Therefore, the area of the sector

= 1/2 × r² × θ

= 1/2 × 14.93² × 3 rad

= 334.35735 m²

Approximately, Area of the sector = 334.4 m²

4 0
2 years ago
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