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AveGali [126]
4 years ago
5

How do u do this ? I am confused on how to do this

Mathematics
1 answer:
Natali5045456 [20]4 years ago
8 0

Answer:

Look to the right. 5 or more? Less? Leave it alone and get rid of the zeros.

Step-by-step explanation:

You might be interested in
Determine consecutive integer values of x between which each real zero is located.
frozen [14]

Answer:

1. x = -2 or x = sqrt(6) - 2 or x = -2 - sqrt(6)

2. x = -2.10947 or x = -0.484343 or x = 1.67884 or x = 2.91497

Step-by-step explanation:

Solve for x:

x^3 + 6 x^2 + 6 x - 4 = 0

The left hand side factors into a product with two terms:

(x + 2) (x^2 + 4 x - 2) = 0

Split into two equations:

x + 2 = 0 or x^2 + 4 x - 2 = 0

Subtract 2 from both sides:

x = -2 or x^2 + 4 x - 2 = 0

Add 2 to both sides:

x = -2 or x^2 + 4 x = 2

Add 4 to both sides:

x = -2 or x^2 + 4 x + 4 = 6

Write the left hand side as a square:

x = -2 or (x + 2)^2 = 6

Take the square root of both sides:

x = -2 or x + 2 = sqrt(6) or x + 2 = -sqrt(6)

Subtract 2 from both sides:

x = -2 or x = sqrt(6) - 2 or x + 2 = -sqrt(6)

Subtract 2 from both sides:

Answer: x = -2 or x = sqrt(6) - 2 or x = -2 - sqrt(6)

_________________________________________

Solve for x:

x^4 - 2 x^3 - 6 x^2 + 8 x + 5 = 0

Eliminate the cubic term by substituting y = x - 1/2:

5 + 8 (y + 1/2) - 6 (y + 1/2)^2 - 2 (y + 1/2)^3 + (y + 1/2)^4 = 0

Expand out terms of the left hand side:

y^4 - (15 y^2)/2 + y + 117/16 = 0

Subtract -3/2 sqrt(13) y^2 - (15 y^2)/2 + y from both sides:

y^4 + (3 sqrt(13) y^2)/2 + 117/16 = (3 sqrt(13) y^2)/2 + (15 y^2)/2 - y

y^4 + (3 sqrt(13) y^2)/2 + 117/16 = (y^2 + (3 sqrt(13))/4)^2:

(y^2 + (3 sqrt(13))/4)^2 = (3 sqrt(13) y^2)/2 + (15 y^2)/2 - y

Add 2 (y^2 + (3 sqrt(13))/4) λ + λ^2 to both sides:

(y^2 + (3 sqrt(13))/4)^2 + 2 λ (y^2 + (3 sqrt(13))/4) + λ^2 = -y + (3 sqrt(13) y^2)/2 + (15 y^2)/2 + 2 λ (y^2 + (3 sqrt(13))/4) + λ^2

(y^2 + (3 sqrt(13))/4)^2 + 2 λ (y^2 + (3 sqrt(13))/4) + λ^2 = (y^2 + (3 sqrt(13))/4 + λ)^2:

(y^2 + (3 sqrt(13))/4 + λ)^2 = -y + (3 sqrt(13) y^2)/2 + (15 y^2)/2 + 2 λ (y^2 + (3 sqrt(13))/4) + λ^2

-y + (3 sqrt(13) y^2)/2 + (15 y^2)/2 + 2 λ (y^2 + (3 sqrt(13))/4) + λ^2 = (2 λ + 15/2 + (3 sqrt(13))/2) y^2 - y + (3 sqrt(13) λ)/2 + λ^2:

(y^2 + (3 sqrt(13))/4 + λ)^2 = y^2 (2 λ + 15/2 + (3 sqrt(13))/2) - y + (3 sqrt(13) λ)/2 + λ^2

Complete the square on the right hand side:

(y^2 + (3 sqrt(13))/4 + λ)^2 = (y sqrt(2 λ + 15/2 + (3 sqrt(13))/2) - 1/(2 sqrt(2 λ + 15/2 + (3 sqrt(13))/2)))^2 + (4 (2 λ + 15/2 + (3 sqrt(13))/2) (λ^2 + (3 sqrt(13) λ)/2) - 1)/(4 (2 λ + 15/2 + (3 sqrt(13))/2))

To express the right hand side as a square, find a value of λ such that the last term is 0.

This means 4 (2 λ + 15/2 + (3 sqrt(13))/2) (λ^2 + (3 sqrt(13) λ)/2) - 1 = 8 λ^3 + 18 sqrt(13) λ^2 + 30 λ^2 + 45 sqrt(13) λ + 117 λ - 1 = 0.

Thus the root λ = 1/4 (-3 sqrt(13) - 5) + (2 2^(2/3) (i sqrt(3) + 1))/(i sqrt(183) - 29)^(1/3) + ((-i sqrt(3) + 1) (i sqrt(183) - 29)^(1/3))/(2 2^(2/3)) allows the right hand side to be expressed as a square.

(This value will be substituted later):

(y^2 + (3 sqrt(13))/4 + λ)^2 = (y sqrt(2 λ + 15/2 + (3 sqrt(13))/2) - 1/(2 sqrt(2 λ + 15/2 + (3 sqrt(13))/2)))^2

Take the square root of both sides:

y^2 + (3 sqrt(13))/4 + λ = y sqrt(2 λ + 15/2 + (3 sqrt(13))/2) - 1/(2 sqrt(2 λ + 15/2 + (3 sqrt(13))/2)) or y^2 + (3 sqrt(13))/4 + λ = -y sqrt(2 λ + 15/2 + (3 sqrt(13))/2) + 1/(2 sqrt(2 λ + 15/2 + (3 sqrt(13))/2))

Solve using the quadratic formula:

y = 1/4 (sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)) + sqrt(2) sqrt((108 - 24 sqrt(13) λ - 16 λ^2 - 4 sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)))/(4 λ + 15 + 3 sqrt(13)))) or y = 1/4 (sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)) - sqrt(2) sqrt((108 - 24 sqrt(13) λ - 16 λ^2 - 4 sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)))/(4 λ + 15 + 3 sqrt(13)))) or y = 1/4 (sqrt(2) sqrt((108 - 24 sqrt(13) λ - 16 λ^2 + 4 sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)))/(4 λ + 15 + 3 sqrt(13))) - sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13))) or y = 1/4 (-sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)) - sqrt(2) sqrt((108 - 24 sqrt(13) λ - 16 λ^2 + 4 sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)))/(4 λ + 15 + 3 sqrt(13)))) where λ = 1/4 (-3 sqrt(13) - 5) + (2 2^(2/3) (i sqrt(3) + 1))/(i sqrt(183) - 29)^(1/3) + ((-i sqrt(3) + 1) (i sqrt(183) - 29)^(1/3))/(2 2^(2/3))

Substitute λ = 1/4 (-3 sqrt(13) - 5) + (2 2^(2/3) (i sqrt(3) + 1))/(i sqrt(183) - 29)^(1/3) + ((-i sqrt(3) + 1) (i sqrt(183) - 29)^(1/3))/(2 2^(2/3)) and approximate:

y = -2.60947 or y = -0.984343 or y = 1.17884 or y = 2.41497

Substitute back for y = x - 1/2:

x - 1/2 = -2.60947 or y = -0.984343 or y = 1.17884 or y = 2.41497

Add 1/2 to both sides:

x = -2.10947 or y = -0.984343 or y = 1.17884 or y = 2.41497

Substitute back for y = x - 1/2:

x = -2.10947 or x - 1/2 = -0.984343 or y = 1.17884 or y = 2.41497

Add 1/2 to both sides:

x = -2.10947 or x = -0.484343 or y = 1.17884 or y = 2.41497

Substitute back for y = x - 1/2:

x = -2.10947 or x = -0.484343 or x - 1/2 = 1.17884 or y = 2.41497

Add 1/2 to both sides:

x = -2.10947 or x = -0.484343 or x = 1.67884 or y = 2.41497

Substitute back for y = x - 1/2:

x = -2.10947 or x = -0.484343 or x = 1.67884 or x - 1/2 = 2.41497

Add 1/2 to both sides:

Answer: x = -2.10947 or x = -0.484343 or x = 1.67884 or x = 2.91497

8 0
3 years ago
Show that the function of f(x) = a - x is its own inverse for all real numbers a
tatuchka [14]

f(x) = a - x

y = a - x

y + x = a - x + x

y + x = a

y + x - y = a - y

x = a - y

f-¹(x) = a - x

5 0
3 years ago
Given that f(x) = x2 + 4x - 32 and g(x) = x– 4, find (f + g)(x) and express
bagirrra123 [75]

Answer:

(f + g)(x) = x² + 5x - 36

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Algebra I</u>

  • Terms/Coefficients/Degrees
  • Function Notation

Step-by-step explanation:

<u>Step 1: Define</u>

f(x) = x² + 4x - 32

g(x) = x - 4

(f + g)(x) is f(x) + g(x)

<u>Step 2: Find</u>

  1. Substitute:                             (f + g)(x) = x² + 4x - 32 + (x - 4)
  2. Combine like terms (x):        (f + g)(x) = x² + 5x - 32 - 4
  3. Combine like terms (Z):        (f + g)(x) = x² + 5x - 36
4 0
3 years ago
Helppppppp please and thanks
Gre4nikov [31]

Answer:

you do not get a answer because I don't know it

3 0
3 years ago
Which statement is true?
Zarrin [17]
This is a problem that has to do with absolute value. Absolute value means the distance from that number to 0, on a number line.

a: |-8|<|8|
The distance from both -8 and 8 to zero, is positive 8 steps. So |-8|=8, as well as 
|8|=8.
8<8 is incorrect.

c: |-7|=-7
7=-7
This is also incorrect.

d: |12|>|-15|
12>15
This inequality is wrong.

Therefore, the only right equation is choice B.

|-2|>1
2>1

This inequality is correct. Two is larger than one.

8 0
3 years ago
Read 2 more answers
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