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garik1379 [7]
2 years ago
5

What is a solutions of x^2-2x+10=0?​

Mathematics
1 answer:
Debora [2.8K]2 years ago
8 0

Answer:

x = 1± 3i

Step-by-step explanation:

x^2-2x+10=0

We can complete the square to solve by subtracting 10 from each side

x^2-2x+10-10=-10

x^2 -2x = -10

We need to add (2/2) ^2  to each side or 1

x^2 -2x+1 = -10 +1

x^2 -2x+1 = -9

The left side factors into (x- (2/2) ) ^2

(x-1) ^2 = -9

Take the square root of each side

sqrt((x-1) ^2 =± sqrt(-9)

x-1 = ±sqrt(-1) sqrt(3)

Remember the sqrt(-1) = i

x-1 = ± 3i

Add 1 to each side

x-1+1 = 1± 3i

x = 1± 3i

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-6z+(-5.5)+3.5z+5y-2.5 simplyfied?
frosja888 [35]

Answer:

  • \boxed{\sf -2.5z+5y-8}

Step-by-step explanation:

\sf -6z+(-5.5)+3.5z+5y-2.5

\sf -6z-5.5+3.5z+5y-2.5

\sf -6z+3.5z+5y-5.5-2.5

\sf( -6z+3.5z) \sf (5y) \sf (-5.5-2.5)

\sf -2.5z+5y-8

\underline{--------------}

Hope it helps! :)

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∠a and ∠e are ______ angles.
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Find the value of a - b when a = 3 and b = -2.
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5

Step-by-step explanation:

a = 3; b = -2

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2 years ago
Find the area of a circle with the radius 9m​
Anuta_ua [19.1K]

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254.47

Step-by-step explanation:

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3 years ago
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See the file below to see my question :)
RideAnS [48]

9514 1404 393

Answer:

  • a = 3
  • b = -18

Step-by-step explanation:

Using the least common denominator, we have ...

  2-\dfrac{x+1}{x-2}-\dfrac{x-4}{x+2}\\\\=\dfrac{2(x-2)(x+2)-(x+1)(x+2)-(x-4)(x-2)}{(x-2)(x+2)}\\\\=\dfrac{2(x^2-4)-(x^2+3x+2)-(x^2-6x+8)}{x^2-4}\\\\=\dfrac{(2-1-1)x^2+(-3+6)x+(-8-2-8)}{x^2-4}=\boxed{\dfrac{3x-18}{x^2-4}}

The values of a and b are ...

  a = 3

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3 years ago
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