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Svetllana [295]
3 years ago
7

-cz+6z= tz+83 solve for z

Mathematics
1 answer:
jenyasd209 [6]3 years ago
7 0

Answer:

z =  83  /( -c+6-t)

Step-by-step explanation:

-cz+6z= tz+83

Subtract tz from each side

-cz+6z -tz= tz-tz+83

-cz+6z- tz=83

Factor out a z

z( -c+6-t) = 83

Divide by ( -c+6-t)

z( -c+6-t)/( -c+6-t) = 83  /( -c+6-t)

z =  83  /( -c+6-t)

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M is a degree 3 polynomial with m ( 0 ) = 53.12 and zeros − 4 and 4 i . Find an equation for m with only real coefficients (i.E.
Nitella [24]

Answer:

Therefore the required polynomial is

M(x)=0.83(x³+4x²+16x+64)

Step-by-step explanation:

Given that M is a polynomial of degree 3.

So, it has three zeros.

Let the polynomial be

M(x) =a(x-p)(x-q)(x-r)

The two zeros of the polynomial are -4 and 4i.

Since 4i is a complex number. Then the conjugate of 4i is also a zero of the polynomial i.e -4i.

Then,

M(x)= a{x-(-4)}(x-4i){x-(-4i)}

      =a(x+4)(x-4i)(x+4i)

      =a(x+4){x²-(4i)²}      [ applying the formula (a+b)(a-b)=a²-b²]

      =a(x+4)(x²-16i²)

      =a(x+4)(x²+16)      [∵i² = -1]

      =a(x³+4x²+16x+64)

Again given that M(0)= 53.12 . Putting x=0 in the polynomial

53.12 =a(0+4.0+16.0+64)

\Rightarrow a = \frac{53.12}{64}

      =0.83

Therefore the required polynomial is

M(x)=0.83(x³+4x²+16x+64)

5 0
3 years ago
Solve the system of linear equations.
sweet-ann [11.9K]

Answer:

  • dependent system
  • x = 2 -a
  • y = 1 +a
  • z = a

Step-by-step explanation:

Let's solve this by eliminating z, then we'll go from there.

Add 6 times the second equation to the first.

  (3x -3y +6z) +6(x +2y -z) = (3) +6(4)

  9x +9y = 27 . . . simplify

  x + y = 3 . . . . . . divide by 9 [eq4]

Add 13 times the second equation to the third.

  (5x -8y +13z) +13(x +2y -z) = (2) +13(4)

  18x +18y = 54

  x + y = 3 . . . . . . divide by 18 [eq5]

Equations [eq4] and [eq5] are identical. This tells us the system is dependent, and has an infinite number of solutions. We can find them in terms of z:

  y = 3 -x . . . . solve eq5 for y

  x +2(3 -x) -z = 4 . . . . substitute into the second equation

  -x +6 -z = 4

  x = 2 - z . . . . . . add x-4

  y = 3 -(2 -z)

  y = z +1

So far, we have written the solutions in terms of z. If we use the parameter "a", we can write the solutions as ...

  x = 2 -a

  y = 1 +a

  z = a

_____

<em>Check</em>

First equation:

  3(2-a) -3(a+1) +6a = 3

  6 -3a -3a -3 +6a = 3 . . . true

Second equation:

  (2-a) +2(a+1) -a = 4

  2 -a +2a +2 -a = 4 . . . true

Third equation:

  5(2-a) -8(a+1) +13a = 2

  10 -5a -8a -8 +13a = 2 . . . true

Our solution checks algebraically.

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2 years ago
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7 0
2 years ago
41°F equals how many degrees Celsius?
GrogVix [38]
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3 0
3 years ago
Read 2 more answers
5. Graph the given system of linear inequalities on the coordinate plane below. (2 points)sy &lt;-3x - 4&lt; 21-15NA11 2 3 4 5 6
xxTIMURxx [149]

ANSWERS

(5) Graph:

(6) Solutions: (-1, -5) and (-2, -7)

Not solutions: (1, 1) and (-4, -3)

EXPLANATION

(5) To graph this system, first, we have to graph each of the inequalities. Both are linear, so we have to graph each line with a dashed line - this is because they are "less than" the line, so the line is not included, and shade the area below each of them.

The first inequality is y < -3x-4,

The second inequality is y < 2x - 1,

The graph of the system is the area shared by both graphs,

(6) Any point that is inside the shaded area is a solution of the system. For example, points (-1, -5) and (-2,-7) are solutions to the system.

On the other hand, any point outside that area, or on the dashed lines is not a solution. For example, points (1, 1) and (-4, -3) are not solutions

8 0
10 months ago
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