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Klio2033 [76]
3 years ago
8

Consider a function

7D" id="TexFormula1" title="f(x) = \frac{(1 - x)^2}{(1 + x)^2}" alt="f(x) = \frac{(1 - x)^2}{(1 + x)^2}" align="absmiddle" class="latex-formula">
Find if a given function is even odd or neither.
Mathematics
1 answer:
SCORPION-xisa [38]3 years ago
8 0

Answer:

f=

x2−2x+1

x3+2x2+x

Step-by-step explanation:

Let's solve for f.

fx=

(1−x)2

(1+x)2

Step 1: Multiply both sides by x^2+2x+1.

fx3+2fx2+fx=x2−2x+1

Step 2: Factor out variable f.

f(x3+2x2+x)=x2−2x+1

Step 3: Divide both sides by x^3+2x^2+x.

f(x3+2x2+x)

x3+2x2+x

=

x2−2x+1

x3+2x2+x

f=

x2−2x+1

x3+2x2+x

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Factorise x ^ 4 + 1/x ^ 4 - 3​
kondaur [170]

Answer:

Step-by-step explanation:

Add 1 and subtract 1, so that the expression won't change.

x^{4}+\frac{1}{x^{4}}-3=x^{4}+\frac{1}{x^{4}}-3 +1 - 1\\\\=x^{4}+\frac{1}{x^{4}} - 2 + 1\\\\= (x^{2})^{2} + (\frac{1}{x^{2}})^{2}- 2 *x^{2}*\frac{1}{x^{2}} - 1\\\\

= (x^{2} -\frac{1}{x^{2}})^{2} - 1\\\\=(x^{2}-\frac{1}{x^{2}})^{2}- 1^{2}              

(a² - b²) = (a+b)(a - b)

Here a = x² - (1/x²)  and b = 1

= [ (x^{2}-\frac{1}{x^{2}})+ 1 ] [ (x^{2} -\frac{1}{x^{2}} - 1 ]

7 0
2 years ago
The number of portable widgets produced weekly by Widget, Inc., is related to the weekly profit in the way:
mixer [17]

9514 1404 393

Answer:

  • 22 widgets
  • $58,400

Step-by-step explanation:

The turning point (maximum or minimum) of quadratic function ...

  f(x) = ax² +bx +c

is at x = -b/(2a).

Here, this means the maximum profit will be had when the number of widgets produced weekly is ...

  x = -88/(2(-2)) = 22

The profit at that level of production is ...

  P(22) = (-2·22 +88)(22) -384 = 44(22) -384 = 968 -384 = 584 . . . hundreds

22 widgets must be produced weekly for the maximum weekly profit of $58,400.

3 0
3 years ago
PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!! PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!! PLSSS HELPPPP I WILLL GIVE YOU B
ozzi

Answer:

72

Step-by-step explanation:

Just multiply all them

7 0
2 years ago
What is the approximate length of the hypotenuse of a right triangle with vertices (-6, 2),
muminat

Answer:

Step-by-step explanation:

A(-6, 2),   B(6,-3) and C(-6, -3)

AB² =  (x₂ - x₁)² + (y₂ - y₁)²

      = ( 6 -[-6])² + ( -3 -2)²

      = ( 6 + 6)² + ( -3 -2)² = 12²  + (-5)²  = 144 + 25 =169

AB = √169 = 13 units

BC²  = ( -6 -6)²  + ( -3 - [-3])² = (-6-6)²  + (-3 +3)²

      = (-12)²  + 0  = 144

BC = √144 = 12 unis

CA²  = (-6 - [-6])²  +(-3-2)²  = (-6 + 6)²  + ( -3-2)²

         = 0 + (-5)²  = 25

CA =√25 = 5 units

length of the hypotenuse of a right triangle  = 13units

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