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posledela
3 years ago
15

Mia is touring New York City with her family. They want to visit the Empire State Building, Central Park, and Times Square befor

e their dinner reservations at 8:45 P.M. They want to spend 1 hour and 15 minutes at the Empire State Building, 2 hours and 30 minutes in Central Park, and 1 hour and 15 minutes in Times Square. What is the latest time Mia's family can start their tour in order to make it to dinner on time?​
Mathematics
1 answer:
Anna [14]3 years ago
3 0

Answer:

3:45pm

Step-by-step explanation:

the reason behind my answer is that in adding all the time they wanted to spend doing things around NYC it totaled 5 hours. 5 hours before 8:45pm is 3:45pm.

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Step-by-step explanation:

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Use the Gram-Schmidt process to find an orthonormal basis for the subspace of R4 spanned by the vectors (1, 0, 1, 1), (1, 0, 1,
Sunny_sXe [5.5K]

Answer:

$ e_1 = \begin{pmatrix}\frac{\sqrt{\textbf{3}}}{\textbf{2}}\\\\\textbf{0} \\\\\frac{\sqrt{\textbf{3}}}{\textbf{3}}\\\\\frac{\sqrt{\textbf{3}}}{\textbf{3}}\end{pmatrix}         $ e_2 = \begin{pmatrix}\frac{\sqrt{\textbf{6}}}{\textbf{6}}\\\\\textbf{0} \\\\\frac{\sqrt{\textbf{6}}}{\textbf{6}}\\\\\frac{\sqrt{\textbf{-6}}}{\textbf{3}}\end{pmatrix}        $ e_3 = \begin{pmatrix}\frac{\sqrt{\textbf{-2}}}{\textbf{2}}\\\\\textbf{0} \\\\\frac{\sqrt{\textbf{2}}}{\textbf{2}}\\\\0\end{pmatrix}

Step-by-step explanation:

we have to orthonormalize the vectors:

v_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \\ 1 \end{pmatrix}    v_2 = \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix}      $ v_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 1 \end{pmatrix}

According to Gram - Schmidt process, we have:

u_k = v_k - \sum_{j = 1} ^ {k - 1} proj_{uj} (v_k) where, $ proj_u (v) = \frac{u . v}{u . u}u

The normalized vector is: $ e_k = \frac{u_k}{\sqrt{u_k.u_k}} $

Now, the first step.

v_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \\ 1 \end{pmatrix} = u₁

Therefore, e₁ = $ \frac{u_1}{\sqrt{u_1.u_1}} $

$  = \begin{pmatrix}\frac{\sqrt{\textbf{3}}}{\textbf{2}}\\\\\textbf{0} \\\\\frac{\sqrt{\textbf{3}}}{\textbf{3}}\\\\\frac{\sqrt{\textbf{3}}}{\textbf{3}}\end{pmatrix}

Now, we find e₂.

$ u_2 = v_2 - \frac{u_1.v_2}{u_1.u_1}u_1 $

$ = \begin{pmatrix} \frac{1}{3}\\ \\ 0 \\\\ \frac{1}{3}\\\\ \frac{-2}{3}  \end{pmatrix}

Therefore, $ e_2 = \frac{u_2}{\sqrt{u_2.u_2}} $

$ e_2 = \begin{pmatrix}\frac{\sqrt{\textbf{6}}}{\textbf{6}}\\\\\textbf{0} \\\\\frac{\sqrt{\textbf{6}}}{\textbf{6}}\\\\\frac{\sqrt{\textbf{-6}}}{\textbf{3}}\end{pmatrix}

To find e₃:

$ u_3 = v_3 - \frac{u_1. v_3}{u_1.u_1}u_1 - \frac{u_2. v_3}{u_2.u_2} u_2 $

$ = \begin{pmatrix} \frac{-1}{2} \\\\ 0\\ \\ \frac{1}{2} \\\\ 0 \\\end{pmatrix}

$ e_3 = \frac{u_3}{\sqrt{u_3.u_3}} $

$ e_3 = \begin{pmatrix}\frac{\sqrt{\textbf{-2}}}{\textbf{2}}\\\\\textbf{0} \\\\\frac{\sqrt{\textbf{2}}}{\textbf{2}}\\\\0\end{pmatrix}

So, we have the orthonormalized vectors $ e_1, e_2, e_3 $.

Hence, the answer.

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Step-by-step explanation:

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