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Fynjy0 [20]
3 years ago
5

G. r. a. p. h. 24x-6y=18.

Mathematics
1 answer:
FinnZ [79.3K]3 years ago
6 0
24x−6y=18 Step 1: Add 6y to both sides. 24x−6y+6y=18+6y24x=6y+18

Step 2: Divide both sides by 24.24x24=6y+1824x=14y+34

Answer: x=14y+34
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In Problems 23–30, use the given zero to find the remaining zeros of each function
Talja [164]

Answer:

x =  2i, x = -2i and x = 4 are the roots of given polynomial.

Step-by-step explanation:

We are given the following expression in the question:

f(x) = x^3 - 4x^2+ 4x - 16

One of the zeroes of the above polynomial is 2i, that is :

f(x) = x^3 - 4x^2+ 4x - 16\\f(2i) = (2i)^3 - 4(2i)^2+ 4(2i) - 16\\= -8i+ 16+8i-16 = 0

Thus, we can write

(x-2i)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16

Now, we check if -2i is a root of the given polynomial:

f(x) = x^3 - 4x^2+ 4x - 16\\f(-2i) = (-2i)^3 - 4(-2i)^2+ 4(-2i) - 16\\= 8i+ 16-8i-16 = 0

Thus, we can write

(x+2i)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16

Therefore,

(x-2i)(x+2i)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16\\(x^2 + 4)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16

Dividing the given polynomial:

\displaystyle\frac{x^3 - 4x^2 + 4x - 16}{x^2+4} = x -4

Thus,

(x-4)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16

X = 4 is a root of the given polynomial.

f(x) = x^3 - 4x^2+ 4x - 16\\f(4) = (4)^3 - 4(4)^2+ 4(4) - 16\\= 64-64+16-16 = 0

Thus, 2i, -2i and 4 are the roots of given polynomial.

4 0
3 years ago
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Mandarinka [93]
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5 0
4 years ago
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In-s [12.5K]

Answer:

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7 0
2 years ago
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Alex777 [14]

it literally gave you the equation lol

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7 0
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professor190 [17]

Answer:

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Step-by-step explanation:

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