Mathematics, 29.10.2020 05:30 reagriffis24

How does a domain restriction placed on a non-invertible function affect its inverse? Drag a function or an interval into each box to correctly complete the statement.

How does a domain restriction placed on a non-invertible function affect its inverse?

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Answers: 3

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Answer from: Quest

answer: plz give a pic and answer choices so i can answer!

step-by-step explanation:

Answer from: Quest

x= 1

step-by-step explanation: convert 5\frac{11}{12}5

to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

ac+b

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{5\times 12+11}{12}3.25+0.5x+1

5×12+11

2 simplify 5\times 125×12 to 6060

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{60+11}{12}3.25+0.5x+1

60+11

3 simplify 60+1160+11 to 7171

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{71}{12}3.25+0.5x+1

4 simplify \frac{5}{6}x

x to \frac{5x}{6}

3.25+0.5x+1\frac{1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+1

5 convert 1\frac{1}{3}1

to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

ac+b

3.25+0.5x+\frac{1\times 3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

1×3+1

6 simplify 1\times 31×3 to 33

3.25+0.5x+\frac{3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

3+1

7 simplify 3+13+1 to 44

3.25+0.5x+\frac{4}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

8 simplify 3.25+0.5x+\frac{4}{3}+\frac{5x}{6}3.25+0.5x+

to \frac{13.75}{3}+\frac{4x}{3}

13.75

\frac{13.75}{3}+\frac{4x}{3}=\frac{71}{12}

13.75

9 simplify \frac{13.75}{3}

13.75

to 4.5833334.583333

4.583333+\frac{4x}{3}=\frac{71}{12}4.583333+

10 subtract 4.5833334.583333 from both sides

\frac{4x}{3}=\frac{71}{12}-4.583333

−4.583333

11 simplify \frac{71}{12}-4.583333

−4.583333 to \frac{4}{3}

\frac{4x}{3}=\frac{4}{3}

12 multiply both sides by 33

4x=\frac{4}{3}\times 34x=

×3

13 simplify \frac{4}{3}\times 3

×3 to \frac{12}{3}

4x=\frac{12}{3}4x=

14 simplify \frac{12}{3}

to 44

4x=44x=4

15 divide both sides by 44

x=1

Answer from: Quest

inverse property is the answer

step-by-step explanation:

Answer from: Quest

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

p1. 1 is in n.

p2. if x is in n, then its "successor" x' is in n.

p3. there is no x such that x' = 1.

p4. if x isn't 1, then there is a y in n such that y' = x.

p5. if s is a subset of n, 1 is in s, and the implication

(x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

def: let a and b be in n. if b = 1, then define a + b = a'

(using p1 and p2). if b isn't 1, then let c' = b, with c in n

(using p4), and define a + b = (a + c)'.

then you have to define 2:

def: 2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem: 1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

then 1 + 1 = 1' = 2 q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

def: let a and b be in n. if b = 0, then define a + b = a.

if b isn't 0, then let c' = b, with c in n, and define

a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

1 + 1 = (1 + 0)'

now use the first part of the definition of + on the sum in

parentheses: 1 + 1 = (1)' = 1' = 2 q.e.d.

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