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Ede4ka [16]
3 years ago
5

WILL GIVE BRAINLIEST!!! Kayla created the graph below to represent the solution to the inequality Negative 8 less-than x.

Mathematics
2 answers:
77julia77 [94]3 years ago
4 0

Answer:

1. she should've pointed the arrow to the right 2.She should have used an open circle at the end of the ray.

Step-by-step explanation:

Nina [5.8K]3 years ago
3 0

_______________________________________________

A and B

________________________________________________

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Use functional notations to describe the function displayed to the right.
LekaFEV [45]

Examining  the table the functional notation is in the second option which is f ( x ) = x^2

<h3>How to determine the  functional notations of the table</h3>

The table represents functions of input variables and the output variables.

The functional notation represents the relationship between the input and out variables

comparing the values and the function

f ( x ) = x^2

at x = -3 ⇒ f ( -3 ) = (-3)^2 = 9

at x = 0 ⇒ f ( 0 ) = (0)^2 = 0

at x = 1 ⇒ f ( 1 ) = (1)^2 = 1

This shows that f ( x ) = x^2 is the correct option

Learn more on functional notation :

brainly.com/question/16779099

#SPJ1

8 0
1 year ago
Express the following relations in the set builder notation. Then, determine whether it is reflexive, symmetric, transitive. Ple
pshichka [43]

Answer:

a)Reflexive, not symmetric, transitive

b)Reflexive, not symmetric, transitive

c)Not reflexive, symmetric, not transitive

d)Reflexive, not symmetric, transitive

Step-by-step explanation:

a)

R=\left \{ (a,b)\epsilon  \mathbb{R} \times \mathbb{R} \mid a \leq b\right \}

The relation R is reflexive for

a\leq a for every real number a

it is not symmetric because 0 is less than 1, but 1 is not less than 0

it is transitive

a\leq and b\leq c\Rightarrow a\leq c

So if aRb and bRc, then aRc

b)  

R=\left \{ (m,n)\epsilon  \mathbb{Z} \times \mathbb{Z} \mid \exists k\in \mathbb{Z} \ni m=kn \right \}

R is reflexive because m=1.m for every integer m

R is not symmetric: 2 is a factor of 4, but 4 is not a factor of 2

R is transitive:  if mRn and nRp if m=kn and n=qp, so m=(kq)p and kq is an integer , so mRp

c)

R=\left \{ (m,n)\epsilon  \mathbb{Z} \times \mathbb{Z} \mid m\neq n\right \}

R is obviously not reflexive because all numbers equals themselves

R is symmetric: if a different to b, then b different to a

R is not transitive: 1R2 and 2R1 (because 1 different to 2), but 1 = 1

d)

R=\left \{ A,B\mid A\subseteq B \right \}

R is reflexive for every set A is a subset of itself

R is not symmetric {1,2} is a subset of {1,2,3} but {1,2,3} is not a subset of {1,2}

R is transitive: if A is subset of B and B is subset of C, then A is subset of C

8 0
4 years ago
What is the LCM of 6 8 and 11
emmainna [20.7K]
The LCM of 6, 8, and 11 is 3 x 2 x 4 x 11, which is 264
6 0
4 years ago
Witch statement is not true?
Valentin [98]
I believe it is A). Please tell me if I'm wrong
7 0
3 years ago
Read 2 more answers
H + 12=-12 H= solve.
babymother [125]
H+12=-12H
12=-13H
12/-13=H
Hope this helps
4 0
3 years ago
Read 2 more answers
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