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STatiana [176]
3 years ago
8

Fina a polynomial function of degree 3 with real coefficients that has -1,2,-4 as zeros

Mathematics
1 answer:
leonid [27]3 years ago
3 0

Answer:

f(x) = x³ + 3x² - 6x - 8

Step-by-step explanation:

given the zeros of a polynomial x = a, x = b, x = c then

(x - a), (x - b), (x - c) are the factors of the polynomial and f(x) is the product of the factors

here x = - 1, x = 2 and x = - 4 are the zeros, hence

(x + 1), (x - 2) and (x + 4) are the factors

f(x) = a(x + 1)(x - 2)(x + 4) ← a is a multiplier

let a = 1, then

f(x) = (x + 1)(x - 2)(x + 4)

     = (x² - x - 2)(x + 4)

     = x³ - x² - 2x + 4x² - 4x - 8

     = x³ + 3x² - 6x - 8


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4/7m = 2/7( 2m+1 )<br> What is M?<br> Check Solution
aev [14]
4/7m = 2/7(2m + 1)
4/7m = 4/7m + 2/7
4/7m - 4/7m = 2/7
0 = 2/7 (incorrect)

no solution
5 0
3 years ago
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John and 3 friends are going out for pizza for lunch. They split one pizza and 4 large drinks. The pizza cost $12.50. They spend
gladu [14]
The answer is $1.25 a large drink

Because 17.5 - 12.5 = 5 / 4 = $1.25


I hope this helps
8 0
3 years ago
Can someone please help me ASAP
Natasha2012 [34]
7^2 +6 ^2= 85
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3 0
3 years ago
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Write an equation of the circle that has a diameter with endpoints (12, 3) and<br> (-18,3).
Mademuasel [1]

The equation of the circle that has a diameter with endpoints (12, 3) and(-18,3) is x² + y² + 6x - 6y - 207 = 0

<h3>Coordinates of center of circle</h3>

Since the endpoints of the diameter are (x₁, y₁) = (12,3) and (x₂, y₂) = (-18,3), the coordinates of the center of the circle are the midpoints of the diameter. So, the midpoints are

  • x = (x₁ + x₂)/2 = (12 + (-18))/2 = (12 - 18)/2 = -6/2 = -3 and
  • y = (y₁ + y₂)/2 = (3 + 3)/2 = 6/2 = 3.

So, the coordinates of the center of the circle are (-3, 3)

<h3>The radius of the circle</h3>

The radius of the circle r = √[(x₁ - h)² + (y₁ - k)²] where

  • (x₁, y₁) = coordinates of end of diameter = (12, 3) and
  • (h, k) = coordinates of center of circle = (-3, 3)

So, substituting the values of the variables into r, we have

r = √[(x₁ - h)² + (y₁ - k)²]

r = √[(12 - (-3))² + (3 - 3)²]

r = √[(12 + 3)² + 0²]

r = √[15² + 0²]

r = √15²

r = 15

<h3>The equation of the circle</h3>

The equation of a circle with center (h,k) is given by

(x - h)² + (y - k)² = r² where r = radius of circle.

Substituting the values of the variables into the equation, we have

(x - h)² + (y - k)² = r²

(x - (-3))² + (y - 3)² = 15²

(x + 3)² + (y - 3)² = 15²

x² + 6x + 9 + y² - 6y + 9 = 225

x² + 6x + y² - 6y + 9 + 9 - 225 = 0

x² + y² + 6x - 6y - 207 = 0

The equation of the circle that has a diameter with endpoints (12, 3) and(-18,3) is x² + y² + 6x - 6y - 207 = 0

Learn more about equation of a circle here:

brainly.com/question/18435467

5 0
2 years ago
Simplified to create an equivalent expression<br><br> (2+2w)•4+9(w+3)=
ahrayia [7]

Answer:

17 w  +  35

Step-by-step explanation:

You get this answer by using PEMDAS.

5 0
3 years ago
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