A learning management system (LMS) is a software application for the administration, documentation, tracking, reporting, and delivery of educational courses, training programs.
Within this course we will guide you on how to use LMS.
REPORT ANY ISSUE RELATING TO LMS USAGE HERE
Quail Semester 2019/2020 Academic Session Continuous Assessment
INSTRUCTION: Answer
all questions; Time Allowed: 10 minutes
Step 1:
Links to assignments can always be found in the sections down the center of the Moodle course page. If your instructor has added the Activities block to the course, you can also locate assignments by clicking the Assignment link in the Activities block, which can usually be found on the right side of the Moodle course page.
Step2:
Select the link to the assignment. The Assignment page will open, showing instructions and your submission status (due date, grading status, time remaining, grade, etc.) for this assignment.
Step 3:
Select Add submission. The Assignment submission page will open. Depending on how your instructor set up the assignment, you may have the option to submit file(s), enter text in a text entry box or both.
Step 4:
To enter online text, type directly into the Online text box. Alternately, you can compose your submission in a word processing program and then cut and paste the text into the Moodle text box.
· When pasting text into Moodle, for best results, paste the content as unformatted text.
· For spell checking to work, it must be enabled in your browser. Once enabled, to see spelling suggestions press and hold the Control key on your keyboard and click OR right-click on your mouse (whether you are using a Mac or PC system will determine which operation will work).
Step 5:
To submit a file using drag-and-drop:
· Arrange your computer desktop windows so that you can view both your browser window and your files on your computer (e.g., in an open folder or on your desktop).
· Drag files from your computer and release them on top of the big blue arrow in the Files area.
· Screenshot of Assignment submission page and an arrow indicating dragging and releasing a .doc file from a browser window.(click to enlarge)
· Note: DO NOT drag-and-drop a .gdoc file from a Drive App desktop window into Moodle. Your instructor will not have permission to view the file. Instead, use browse for the file (for details, see the next step), or save the file as a PDF before submitting.
Step 6:
To browse for a file:
I. Under File submissions, at the top-left of the file area, click the Add... icon (Screenshot of Add File icon, the outline of a paper document with the top right corner folded over.). The File picker window will open.
II. Select a file from your computer, or from Google Drive:
· To upload a file from your computer, click Upload a file (at left), then, below Attachment, select Choose file. Browse your computer and select a file, then click Open. The file name will appear next to the Choose file button.
· To upload a file from Google Drive, click Google Drive (at left). You may need to log in. Click the File to upload.
Note: When you browse to Google Drive, Moodle will convert Google docs (.gdoc files) that were composed directly in Google to a sharable file type (.rtf, .ppt, .xls, etc.)
III. (Optional) If you want to rename your file in the File Picker, enter the new name in the Save as field, and make sure to include the correct file extension (.pdf, .doc, etc.) in the name so the file can open correctly for your instructor.
IV. Click Upload this file. The File Picker will close and an icon for your uploaded file will appear in the Files area.
Step 7:
Once your file(s) appear(s) in the Files submission area, click Save changes.
Note: You should receive an email at your umass.edu mail address confirming you have submitted the assignment.
Step 8:
If your instructor allows you to revise your submission, you'll see an Edit submission button on the Assignment page. Select Edit submission to add or replace files, or edit a text submission. When you finish making changes, click Save changes.
Step 9:
For some assignment types, you will see a Submit assignment button. To finalize your submission, click Submit assignment, then Continue.
Note: Once finalized, you will not be able to make any more changes.
Step 10:
You can check under Submission status on the Assignment page to ensure your file is listed. You can also view feedback and grades from your instructor there.
Step 11:
For some assignment types, you may have the option to submit several or an unlimited amount of attempts (or drafts) for a single assignment. Because each course is different, you will need to communicate with your instructors on how submission attempts and feedback are configured for each Moodle assignment.
This will be done using some mathematical question and show easy and fast ways to attend to it.
1. Differentiate \( y=4t^3+6t^2-3t \)
2. Integrate \( dy= \int_{-2}^{2}{9x^2dx} \)
3. Matrices
$$ A= \left[\begin{array}{cc} 2&4\\1&3\end{array}\right] $$
$$B= \left[\begin{array}{cc} 3&1\\-4&2\end{array}\right] $$
Find:
a. A+B
b. 2A-B
c. AxB
d. /A/
4. Using the Regular-Falsi to compute the root for the function
f(x) = x2 -9x+1 using interval (2,3)
1. Solve with pencil and paper first
2. Write the code to the space provided
3. In case of substituting numerical figures to the equation copy the Tex code written in 2 above and change the representing alphabet with respective numerical
4.The code will change after click on finish bottom
So it appear as
$$ y = 4t^3+6t^2-3t $$
\( \frac{dy}{dx}=12t^2+12t-3 \)
\( \frac{d^2y}{dx^2}=24t+12 \)
So it appears as:
$$ dy= \int_{-2}^{2}{9x^2dx} $$
\( y= ({\frac{9x^3}{3}})_{-2}^{2} \)
\( y= ({3x^3})_{-2}^{2} \)
\( y=3(2)^3-3(-2)^3
\)
\( y=3(8)-3(-8)\)
\(y=24+24)\)
$$y=48$$
a. A+B =\( \left[ \begin{array}{cc} 2&4 \\ 1&3 \end{array} \right]+ \left [\begin{array}{cc} 3&1 \\ -4&2 \end{array} \right] \)
$$= \left[ \begin{array}{cc} 2+3&4+1 \\ 1+(-4)&3+2 \end{array} \right]$$
$$= \left[ \begin{array}{cc} 5&5 \\ -3&5 \end{array} \right]$$
b. 2A-B =\( 2\times \left[ \begin{array}{cc} 2&4 \\ 1&3 \end{array} \right]+ \left [\begin{array}{cc} 3&1 \\ -4&2 \end{array} \right] \)
$$=\left [ \begin{array}{cc} 4&8 \\ 2&6 \end{array} \right]- \left [\begin{array}{cc} 3&1 \\ -4&2 \end{array} \right] $$
$$= \left[ \begin{array}{cc} 4-3&8-1 \\ 2-(-4)&6-2 \end{array} \right]$$
$$= \left[ \begin{array}{cc} 1&7 \\ 6&4 \end{array} \right]$$
c. AxB =\( \left[ \begin{array}{cc} 2&4 \\ 1&3 \end{array} \right]\times \left [\begin{array}{cc} 3&1 \\ -4&2 \end{array} \right] \)
$$= \left[ \begin{array}{cc} -10&10 \\ -9&7 \end{array} \right]$$
d. /A/=\( \left[ \begin{array}{cc} 2&4 \\ 1&3 \end{array} \right]\)
$$=(2 \times 3)-(1 \times 4)$$
$$=6-4$$
$$=2$$
The given function is
$$f(x)=x^2-9x+1$$
For interval [2, 3]; When x=2; \(f(2)=(2)^2-9(2)+1= -9 \)
When x=3; \(f(3)=(3)^2-9(3)+1=1\)
Since f(x=2) and f(x=3) gives opposite signs, then there is root in the interval, Hence apply formula:
$$ X_n+_1=X_n- \frac{X_n-X_n-_1}{f(X_n)-f(X_n-_1)}f(X_n) $$
First Iteration: Since X1=2 and X2=3 with respect to f(X1)=-9 and f(X2)=1
Therefore X3 is first determine as first iteration
$$ X_3=X_2- \frac{X_2-X_1}{f(X_2)-f(X_1)}f(X_2) $$
$$ X_3=3- \frac{3-2}{1-(-9)}(1) $$
$$ X_3=3- \frac{1}{10}(1) =3-0.1=2.9 $$
Now Check for f(x=2.9) in the function; \(f(2.9)=(2.9)^2-9(2.9)+1= -0.711 \)
Since the new f(X=2.9) sign is negative it replaces f(X=2) from negative end.
Hence, the new interval is (2.9,3)
Second iteration:
$$ X_4=X_3- \frac{X_3-X_2}{f(X_3)-f(X_2)}f(X_3) $$
$$ X_4=3- \frac{3-2.9}{1-(-0.711)}(1) $$
$$ X_4=3- \frac{0.1}{1.711)}(1)=2.942 $$
Now Check for f(x=2.942) in the function; \(f(2.942)=(2.942)^2-9(2.942)+1= -0.0139 \)
Since the new f(X=2.942) sign is negative it replaces f(X=2.9) from negative end.
Hence, the new interval is (2.942,3)
Third iteration:
$$ X_5=X_4- \frac{X_4-X_3}{f(X_4)-f(X_3)}f(X_4) $$
$$ X_5=3- \frac{3-2.942}{1-(-0.0139)}(1) $$
$$ X_5=3- \frac{0.058}{1.0139)}(1)=2.9428 $$
Now Check for f(x=2.9428) in the function; \(f(2.9428)=(2.9428)^2-9(2.9428)+1= 0.0034\)
Since the new f(X=9428) sign is positive it replaces f(X=3) from positive end.
Hence, the new interval is (2.942, 2.9428)
Fourth iteration:
$$ X_6=X_5- \frac{X_5-X_4}{f(X_5)-f(X_4)}f(X_5) $$
$$ X_6=3- \frac{3-2.9428}{1-(0.0034)}(1) $$
$$ X_6=3- \frac{0.0572}{0.9966)}(1)=2.9426 $$
Now Check for f(x=2.9426) in the function; \(f(2.9426)=(2.9426)^2-9(2.9426)+1= 0.0037\)
Since the new f(X=9426) sign is positive it replaces f(X=3) from positive end.
Hence, the new interval is (2.942, 2.9426). Fifth iteration:
$$ X_7=X_6- \frac{X_6-X_5}{f(X_6)-f(X_5)}f(X_6) $$
$$ X_7=3- \frac{3-2.9426}{1-(0.0037)}(1) $$
$$ X_7=3- \frac{0.0572}{0.9966)}(1)=2.9426 $$
This shows that the root of the equation falls at 2.9426
\( \left| \begin{array}{ccc} 4&-4&21 \\ 15&2&-1\\6&18&0 \end{array} \right| \left| \begin{array}{ccc} I_1 \\ I_2\\I_3 \end{array} \right|= \left| \begin{array}{ccc} 20 \\ 0\\15 \end{array} \right| \)