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Matching (2 question, 10 points)

Your client has severe peripheral arterial disease. When the lower extremities are elevated you would expect them to appear [sign1] and when they are in the dependent position you would expect them to appear [sign2]. Fill in the blanks. 

For the listed drugs below (generic or Trade name), match the classification/classifications from the following list:  (25 pts)

A client is suspected to have Bell’s palsy. What priority nursing intervention should be provided to the client?

I agree to uphold my school’s Academic Code of Integrity and Academic Integrity policy of this course.  I also understand that I must remain in view of webcam for the entire duration of my assessment.

Please download the test questions pdf here: Test1_ques.pdf and return to this test window to upload your answers. Do NOT CLOSE this test window.   

Do you affirm that all work you submit is your own and that you will not consult any other person or resources?

QUESTION 6(a) for Gradescope: If  u  and  v {“version”:”1.1″,”math”:”u \text{ and } v”} are non-zero vectors in n-space, then  span { u , v } {“version”:”1.1″,”math”:”\text{span}\{u,v\}”}does NOT contain the zero vector. 

QUESTION 3 for Gradescope: Consider the linear transformation with matrix representation A=[21430−1].{“version”:”1.1″,”math”:”A=\begin{bmatrix} 2 & 1 & 4 \\ 3 & 0 & -1 \end{bmatrix}.”} Find an intrinsic description of the kernel of this linear transformation. Show all your work. State the nullity and rank of A based on your work for finding the kernel.

QUESTION 2 for Gradescope: (a) Use the Marquis de LaPlace (aka Gram-Schmidt) process to find an orthogonal basis with the same span as { [ 2 4 2 ] , [ 2 1 − 1 ] } {“version”:”1.1″,”math”:”\left\{ \begin{bmatrix} 2 \\ 4\\ 2 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\-1\end{bmatrix} \right\}”} Note that your orthogonal basis does not need to be orthonormal. (b) Use the orthogonal basis found in part (a) to find the least squares solution to the following problem. Do not use a method to find the least squares solution that was not covered in the lectures. {2x+2y=04x+y=−22x−y=1{“version”:”1.1″,”math”:”\begin{cases} 2x+2y & =0 \\ 4x+y & = -2 \\ 2x-y & = 1 \end{cases}”}