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Graph y1 = 2x +1 if -1< x< 0

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k24674

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  • Texas Instru... Master 8,093 Answers
  • Posted on Oct 30, 2010
k24674
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Graph 2X PLUS 1 for X in open interval ]-1,0[

It should be entered as follows
(2X plus 1) (X larger than negative 1) (X less than zero)

In [Y=] editor and on line Y1= type your right hand side between parentheses (2X Plus 1). I use the Plus instead of the usual sign because the parser of the web site removes the sign.
[( ]2 [X,T,theta,n] [Plus] 1 [)] [(] [X,T,theta,n] [2nd][MATH] [3: larger than] [(-)] 1 [)] [(] [X,T,Theta,n] [2nd][MATH][5: less than] 0 [)]
Here are some screen captures to help you
graph y1 = 2x +1 if -1< x< 0 - ecfa24f.jpgfa02graph y1 = 2x +1 if -1< x< 0 - d83c851.jpgass=
d83c851.jpg

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Graph 1 2 on the number line.

https://www.youtube.com/watch?v=FHOE4EvYCm8

https://www.youtube.com/watch?v=IDG20QMlcm8

Graphing on a Number Line
Integers and real numbers can be represented on a number line. The point on this line associated with each number is called the graph of the number. Notice that number lines are spaced equally, or proportionately (see Figure 1).
Figure 1. Number lines.

Graphing inequalities
When graphing inequalities involving only integers, dots are used. Example 1 Graph the set of x such that 1 ? x ? 4 and x is an integer (see Figure 2).
{ x:1 ? x ? 4, x is an integer}

Figure 2. A graph of {x:1 ? x ? 4, x is an integer}.
When graphing inequalities involving real numbers, lines, rays, and dots are used. A dot is used if the number is included. A hollow dot is used if the number is not included.
Example 2 Graph as indicated (see Figure 3).
1.Graph the set of x such that x ? 1.
{ x: x ? 1}
2.Graph the set of x such that x > 1 (see Figure 4).
{ x: x > 1}
3.Graph the set of x such that x < 4 (see Figure 5).
{ x: x < 4}
This ray is often called an open ray or a half line. The hollow dot distinguishes an open ray from a ray.

Figure 3. A graph of { x: x ? 1}.

Figure 4. A graph of { x: x > 1}

Figure 5. A graph of { x: x < 4}
Intervals
An interval consists of all the numbers that lie within two certain boundaries. If the two boundaries, or fixed numbers, are included, then the interval is called a closed interval. If the fixed numbers are not included, then the interval is called an open interval. Example 3 Graph.
1.Closed interval (see Figure 6).
{ x: -1 ? x ? 2}
2.Open interval (see Figure 7).
{ x: -2 < x < 2}

Figure 6. A graph showing closed interval { x: -1 ? x ? 2}.

Figure 7. A graph showing open interval { x: -2 < x < 2}.
If the interval includes only one of the boundaries, then it is called a half-open interval.
Example 4 Graph the half-open interval (see Figure 8).
{ x: -1 < x ? 2}

Figure 8. A graph showing half-open interval { x: -1 < x ? 2}.

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You focused on system errors. I recommend you read the MacOS 10.14 review here - https://osxtips.net/ This will add new opportunities for you to resolve the issue. This method always helps me. Try to follow the instructions in the review, maybe this will remove the current problem.
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Every time I try to graph on TI 84 it says ERR: INVALID DIM 1:QUIT I have enterered something simple y= x+2 and window is set for -10&lt;x&lt;10 and -10&lt;y&lt;10 the curser in y=...

Go to http://education.ti.com/downloads/guidebooks/graphing/84p/TI84Plus_guidebook_EN.pdf to download the manual. On page 401, you will find the following:
INVALID DIM • The ERR:INVALID DIM error message may occur if you are trying to graph a function that does not involve the stat plot features. The error can be corrected by turning off the stat plots. To turn the stat plots off, press y , and then select 4:PlotsOff. • You specified a list dimension as something other than an integer between 1 and 999. • You specified a matrix dimension as something other than an integer between 1 and 99. • You attempted to invert a matrix that is not square.
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7_31_2012_12_55_37_pm.jpg


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How to be solve banker algorithm?example

I m providing you this from my college notes

Banker's Algorithm


* multiple instances of resource types IMPLIES cannot use resource-allocation graph

* banks do not allocate cash unless they can satisfy customer needs when a new process enters the system

* declare in advance maximum need for each resource type

* cannot exceed the total resources of that type

* later, processes make actual request for some resources

* if the the allocation leaves system in safe state grant the resources

* otherwise, suspend process until other processes release enough resources



Banker: Data Structures define MAXN 10 /* maximum number of processes */
#define MAXM 10 /* maximum number of resource types */
int Available[MAXM]; /* Available[j] = current # of unused resource j */
int Max[MAXN][MAXM]; /* Max[i][j] = max demand of i for resource j */
int Allocation[MAXN][MAXM]; /* Allocation[i][j] = i's current allocation of j*/
int Need[MAXN][MAXM]; /* Need[i][j] = i's potential for more j */
/* Need[i][j] = Max[i][j] - Allocation[i][j] */

Notation:

X <= Y iff X[i] <= Y[i] for all i

(0,3,2,1) is less than (1,7,3,2)

(1,7,3,2) is NOT less than (0,8,2,1)

Each row of Allocation and Need are vectors: Allocation_i and Need_i



Banker: Example

Initially:

Available
A B C
10 5 7

Later Snapshot:

Max - Allocation = Need Available
A B C A B C A B C A B C
P0 7 5 3 0 1 0 7 4 3 3 3 2
P1 3 2 2 2 0 0 1 2 2
P2 9 0 2 3 0 2 6 0 0
P3 2 2 2 2 1 1 0 1 1
P4 4 3 3 0 0 2 4 3 1



Banker: Safety Algorithm

* consider some sequence of processes

* if the first process has Need less than Available

* it can run until done

* then release all of its allocated resources

* allocation is increased for next process

* if the second process has Need less than Available

* ...

* then all of the processes will be able to run eventually

* IMPLIES system is in a safe state



Banker: Safety Algorithm

STEP 1: initialize
Work := Available;
for i = 1,2,...,n
Finish[i] = false
STEP 2: find i such that both
a. Finish[i] is false
b. Need_i <= Work
if no such i, goto STEP 4
STEP 3:
Work := Work + Allocation_i
Finish[i] = true
goto STEP 2
STEP 4:
if Finish[i] = true for all i, system is in safe state



Banker: Safety Example

Using the previous example, P1,P3,P4,P2,P0 satisfies criteria.

Max - Allocation = Need <= Work Available
A B C A B C A B C A B C
P1 3 2 2 2 0 0 1 2 2 3 3 2 3 3 2
P3 2 2 2 2 1 1 0 1 1 5 3 2
P4 4 3 3 0 0 2 4 3 1 7 4 3
P2 9 0 2 3 0 2 6 0 0 7 4 5
P0 7 5 3 0 1 0 7 4 3 10 4 7
10 5 7<<< initial system



Banker: Resource-Request Algorithm

STEP 0: P_i makes Request_i for resources, say (1,0,2)
STEP 1: if Request_i <= Need_i
goto STEP 2
else ERROR
STEP 2: if Request_i <= Available
goto STEP 3
else suspend P_i
STEP 3: pretend to allocate requested resources
Available := Available - Request_i
Allocation_i := Allocation_i + Request_i;
Need_i := Need_i - Request_i
STEP 4: if pretend state is SAFE
then do a real allocation and P_i proceeds
else
restore the original state and suspend P_i



Banker: Resource-Request Algorithm [129]

Say P1 requests (1,0,2)

Compare to Need_1: (1,0,2) <= (1,2,2)

Compare to Available: (1,0,2) <= (3 3 2)

Pretend to allocate resources:

Max - Allocation = Need Available
A B C A B C A B C A B C
P0 7 5 3 0 1 0 7 4 3 2 3 0<<<
P1 3 2 2 3 0 2<<< 0 2 0<<<
P2 9 0 2 3 0 2 6 0 0
P3 2 2 2 2 1 1 0 1 1
P4 4 3 3 0 0 2 4 3 1

Is this safe? Yes: P1, P3, P4, P0, P2

Can P4 get (3,3,0)? No, (3,3,0) > (2,3,0) Available

Can P0 get (0,2,0)? (0,2,0) < (2,3,0) Available

Pretend: Available goes to (2,1,0)

Thanks And Regards
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My door locks do not operate when using remote on key fob

If it's the remote buttons wearing out (check by taking remote apart and bridging the contacts with a small screwdriver. If it works fine that way, go to keylessfix.com $10.00 will fix it.
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