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zhannawk [14.2K]
3 years ago
5

Find the zeros of each function. Part 3. Please show work.

Mathematics
2 answers:
Helen [10]3 years ago
7 0

9514 1404 393

Answer:

real: -1, 2; complex: +i, -i

1, 3, 4

Step-by-step explanation:

1. The graph (red) shows the only real zeros to be -1 and 2. When the corresponding factors are divided from the function, the remaining factor is the quadratic (x^2 +1), which has only complex roots. The quadratic is graphed in green.

The linear factorization is ...

 f(x) = (x +1)(x -2)(x -i)(x +i)

The roots are -1, 2, -i, +i.

__

2. The graph (blue) shows the zeros are 1, 3, 4.

You observe that the sum of coefficients is zero, so x=1 is a root. Factoring that out gives the quadratic (x^2 -7x +12), which you recognize factors as

 (x -3)(x -4) . . . zeros of 3 and 4

__ The tables shown correspond to f1(x)/(x-2) and f2(x)/(x-1).

Novay_Z [31]3 years ago
6 0

9514 1404 393

Answer:

  1. real: -1, 2; complex: +i, -i
  2. 1, 3, 4

Step-by-step explanation:

1. The graph (red) shows the only real zeros to be -1 and 2. When the corresponding factors are divided from the function, the remaining factor is the quadratic (x^2 +1), which has only complex roots. The quadratic is graphed in green.

The linear factorization is ...

  f(x) = (x +1)(x -2)(x -i)(x +i)

The roots are -1, 2, -i, +i.

__

2. The graph (blue) shows the zeros are 1, 3, 4.

You observe that the sum of coefficients is zero, so x=1 is a root. Factoring that out gives the quadratic (x^2 -7x +12), which you recognize factors as

  (x -3)(x -4) . . . zeros of 3 and 4

__

I have attached a spreadsheet that does synthetic division. There are web sites that will do this, too. The tables shown correspond to f1(x)/(x-2) and f2(x)/(x-1). When you fill in the zero and coefficients, the built-in formulas do the rest.

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3 years ago
Find the orthogonal complement W⊥ of W and give a basis for W⊥. W = x y z : x = 1 2 t, y = − 1 2 t, z = 6t.
igor_vitrenko [27]

Answer:

W⊥= (a, b, c)

a = v - (1/2)w

b = v

c = w

Basis for W⊥ = <(1,1,0),(-1/2,0,1)>

Step-by-step explanation:

The orthogonal complement of W is the set of vectors (a,b,c) that satisfy:

(x,y,z)·(a,b,c) = 0

ax + by + cz = 0

So, taking into account that x=12t, y=-12t and z=6t, the equation above is equal to:

12ta + -12tb + 6tc = 0

12a - 12b + 6c = 0

Then, if we made b=v and c=w and solve for a, we get:

12a - 12v + 6w = 0

a = (12v - 6w)/12

a = v - (1/2)w

Therefore, the orthogonal complement W⊥ of W has the form:

W⊥ = (a,b,c) = (v - (1/2)w, v, w)

On the other hand, we can write W⊥ as:

W⊥ = (v - (1/2)w, v, w)

W⊥ = (v,v,0) + ((-1/2)w,0,w)

W⊥ = v(1,1,0) + w(-1/2,0,1)

That means that we can write any vector of W⊥ as a linear combinations of the vectors (1,1,0) and (-1/2,0,1), so they are a basis for W⊥

7 0
4 years ago
. Un bloque de hierro tiene 5.0 cm de largo, 3.0 cm de alto y 4.0 cm de ancho y pesa 474 g ¿Cuál es la densidad del hierro?
mrs_skeptik [129]

Answer:

the density of the iron is  7.9 g /cm^3

Step-by-step explanation:

The computation of the density of the iron is shown below:

Density = weight ÷ volume

= 474 ÷ (5 × 3 × 4)

= 474 ÷ 60

= 7.9 g /cm^3

Hence, the density of the iron is  7.9 g /cm^3

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