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lianna [129]
3 years ago
10

Help! write the set of numbers in set-builder notation: the set of all real numbers except 100​

Mathematics
2 answers:
nignag [31]3 years ago
7 0

Answer:

{ x | x ∈ R, x ≠ 100 }

Step-by-step explanation:

Real numbers (R) include all rational and irrational numbers.

Hence, If we need to write it in set-builder notation, we will write as:

{ x | x ∈ R, x ≠ 100 }

<u>In words :</u> "x such that x belongs to R, x is not equal to 100"

This shows that x is any real number except for 100.

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3><h3>Peace!</h3>
k0ka [10]3 years ago
3 0

Answer:

Step-by-step explanation:

The set of all real numbers except for 100 in set builder notation (assuming the variable is x) is:

The very first x represents the variable.

The second with the R represents the number. Here, we want all real numbers.

And the third is the constraint. We want all real numbers except for 100.

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Step-by-step explanation:

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1. What is the vertex form of the equation?
sergeinik [125]

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to get into vertex form, complete the square

y=-x^2+12x-4

step 1: group x terms

y=(-x^2+12x)-4

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y=-1(x^2-12x)-4

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step 3: add positive and negative of that previous number inside the parenthasees

y=-1(x^2-12x+36-36)-4

factor perfect squaer trionmial

y=-1((x-6)^2-36)-4

take out that -36 by expansion

y=-1(x-6)^2+36-4

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2. factor out the 2

2(x^2+8x+12)

what 2 numbers multiply to get 12 and add to get 8?

2 and 6

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3.

for ax^2+bx+c=0

x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}

in our case

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a=1, b=-7, c=-6

x=\frac{-(-7) \pm \sqrt{(-7)^2-4(1)(-6)}}{2(1)}

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4. remember that i=\sqrt{-1} so i^2=-1

treat i as a variable

(-2i)(8i)=

(-2)(i)(8)(i)=

(-2)(8)(i)(i)=

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(-16)(-1)=

[tex}16[/tex]

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