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Slav-nsk [51]
3 years ago
9

Factor the polynomial completely using any method. Show steps. 2d^4-13d^2-45

Mathematics
1 answer:
Alinara [238K]3 years ago
3 0

Answer:

\huge ( {2d}^{2}  + 5)(d + 3)(d - 3)

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>
  • factoring
  • PEMDAS
<h3>given:</h3>
  • 2d^4-13d^2-45
<h3>let's solve:</h3>

2d^4-13d^2-45

2 {d}^{4}  + 5 {d}^{2}  - 18 {d}^{2}  - 45 \\  {d}^{2} ( {2d}^{2}  + 5) - 9( {2d}^{2}  + 5) \\ ( {2d}^{2}  + 5)( {d}^{2}  - 9 )\\ ( {2d}^{2}  + 5)(d + 3)(d - 3)

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What is the area?<br><br> Write your answer as a fraction or as a whole or mixed number.
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Step-by-step explanation:

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3 years ago
Find the GCF<br> 34 &amp; 68
photoshop1234 [79]

Answer:

The GCF would be 34.

34/34=1

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3 years ago
Michelle bought 12 cookies for $2.70. At that rate, how much would it cost her to buy 20 cookies?
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3 years ago
The function f (x comma y )equals 3 xy has an absolute maximum value and absolute minimum value subject to the constraint 3 x sq
zmey [24]

Answer:

The maximum value of f is 363, which is reached in (11,11) and (-11,-11) and the minimum value of f is -33, which is reached in (√11,-√11) and (-√11,√11)

Step-by-step explanation:

f(x,y) = 3xy, lets find the gradient of f. First lets compute the derivate of f in terms of x, thinking of y like a constant.

f_x(x,y) = 3y

In a similar way

f_y(x,y) = 3x

Thus,

\nabla{f} = (3y,3x)

The restriction is given by g(x,y) = 121, with g(x,y) = 3x²+3y²-5xy. The partial derivates of g are

[ŧex] g_x(x,y) = 6x-5y [/tex]

g_y(x,y) = 6y - 5x

Thus,

\nabla g(x,y) = (6x-5y,6y-5x)

For the Langrange multipliers theorem, we have that for an extreme (x0,y0) with the restriction g(x,y) = 121, we have that for certain λ,

  • f_x(x_0,y_0) = \lambda \, g_x(x0,y0)
  • f_y(x_0,y_0) = \lambda \, g_y(x_0,y_0)
  • g(x_0,y_0) = 121

This can be translated into

  • 3y = \lambda (6x-5y)
  • 3x = \lambda (-5x+6y)
  • 3 (x_0)^2 + 3(y_0)^2 - 5\,x_0y_0 = 121

If we sum the first two expressions, we obtain

3x + 3y = \lambda (x+y)

Thus, x = -y or λ=3.

If x were -y, then we can replace x for -y in both equations

3y = -11 λ y

-3y = 11 λ y, and therefore

y = 0, or λ = -3/11.

Note that y cant take the value 0 because, since x = -y, we have that x = y = y, and g(x,y) = 0. Therefore, equation 3 wouldnt hold.

Now, lets suppose that λ=3, if that is the case, we can replace in the first 2 equations obtaining

  • 3y = 3(6x-5y) = 18x -15y

thus, 18y = 18x

y = x

and also,

  • 3x = 3(6y-5x) = 18y-15x

18x = 18y

x = y

Therefore, x = y or x = -y.

If x = -y:

Lets evaluate g in (-y,y) and try to find y

g(-y,y) = 3(-y)² + 3y*2 - 5(-y)y = 11y² = 121

Therefore,

y² = 121/11 = 11

y = √11 or y = -√11

The candidates to extremes are, as a result (√11,-√11), (-√11, √11). In both cases, f(x,y) = 3 √11 (-√11) = -33

If x = y:

g(y,y) = 3y²+3y²-5y² = y² = 121, then y = 11 or y = -11

In both cases f(11,11) = f(-11,-11) = 363.

We conclude that the maximum value of f is 363, which is reached in (11,11) and (-11,-11) and the minimum value of f is -33, which is reached in (√11,-√11) and (-√11,√11)

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4 years ago
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