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kogti [31]
3 years ago
6

For each of the following functions, determine if they are injective. Also determine if theyare surjective. Also determine if th

ey are bijective.No justification needed.The firstthree have already been filled out for you, as an example. Also, recall interval notation:[a,b] ={x∈R:a≤xandx≤b}. You should be able to come up with a similar definitionfor (a,b) and (a,b] and [a,b).1.f:R→R, f(x) =x2.f:R≥0→R, f(x) =x+ 13.f:R≥0→R, f(x) = cos(x)4.f:R→R, f(x) =ex5.f:R→R, f(x) =2xifx≥0−x2ifx <0.6.f:R→(0,[infinity]), f(x) =ex7.f:R≥0→R≥0, f(x) =x48.f:{−1,2,−3}→{1,4,9}, f(x) =x29.f:R≥0→[−1,1], f(x) = cos(x)10.f: [0,π]→[−1,1], f(x) = cos(x)11.f:R≥0→[−1,1], f(x) = 012.f: US Citizens→Z, f(x) = the SSN ofx.13.f: US Zip Codes→US States, f(x) = The state thatxbelongs to.

Mathematics
1 answer:
timurjin [86]3 years ago
7 0

Answer:

Check below

Step-by-step explanation:

1. Definition for intervals

(a,b)=\left \{ x\in\Re : a

2. Functions

1) \Re \rightarrow \Re \\ f(x)=x

Let's perform graph tests.

That's an one to one, injective function. Look how any horizontal line touches that only once. Also, It's a surjective and a bijective one.

2)\Re\geq0\rightarrow\Re , f(x)=x+1\\

Injective, surjective and bijective.

Injective: a horizontal line crosses the graph in one point.

3)f:\Re\geq 0\rightarrow\Re, f(x) = cos(x)

The cosine function is not injective, bijective nor surjective.

4)f:\Re\rightarrow\Re \:f(x)=ex

Since e, is euler number it's a constant. It's also injective, surjective and bijective.

5)  Quite unclear format

6) \:f:\Re\rightarrow(0,\infty), f(x) =ex

Despite the Restriction for the CoDomain, the function remains injective, surjective and therefore bijective.

7) f:\Re\geq 0 \rightarrow  \Re\geq0, f(x) =x^{4}

Not injective nor surjective therefore not bijective too.

8).f:\{-1,2,-3\}}\rightarrow \{1,4,9\}, f(x) =x^{2}

f(-1)=1, f(2)=4, f(-3)=9

Injective (one to one), Surjective,  and Bijective.

10) f:\Re\geq 0\rightarrow [-1,1], f(x)= cos(x)\\-1=cos(x) \therefore x=\pi,3\pi,5\pi,etc.

Surjective.

11.f:R\geq 0[-1,1], f(x) = 0\\

Surjective

12.f: US Citizens→Z, f(x) = the SSN of x.

General function

13.f: US Zip Codes→US States, f(x) = The state that x belongs to.

Surjective

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Answer: Lucy pays $ 492.58  back altogether.

Step-by-step explanation:

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Work out m and c for the line:2y+3x=5
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Step-by-step explanation:

General line equation: y = mx + c.

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3 years ago
Please help me answer this for a test im having trouble
Dimas [21]

Answer:

x + 1

y = 9

Step-by-step explanation:

In order to solve this question we need to represent "y "in terms of "x" in the first equation, and the plug in the "y" value in the first equation into the second one. Luckily for us in the first equation it already shows what "y" is equal to in terms of "x" (based on the first equation y = -x + 10). Now we just need to plug in the value that we got instead of "y" in the second equation, and so we get....

y = 7x + 2

(plug in the y value and get the following ….)

-x + 10 = 7x + 2

(now just solve the following equation)

-x + 10 + x = 7x + 2 + x

10 = 8x + 2

10 - 2 = 8x + 2 - 2

8 = 8x

8/8 = 8x/8

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Now that we know the value of "x", all we need to do now is substitute the value of "x" into any of the equations and we will get the value of "y". So we do the following.....

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3 0
3 years ago
URGENT PLEASE HELP
Inga [223]

Answer:

y = - \frac{1}{3} x + 2

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = 3x - 3 ← is in slope- intercept form

with slope m = 3

Given a line with slope m then the slope of a line perpendicular to it is

m_{perpendicular} = - \frac{1}{m} = - \frac{1}{3}, hence

y = - \frac{1}{3} x + c ← is the partial equation of the perpendicular line

To find c substitute (3, 1) into the partial equation

1 = - 1 + c ⇒ c = 1 + 1 = 2

y = - \frac{1}{3} x + 2 ← equation of perpendicular line

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