Find the square root of these numbers to the nearest tenth 62,23,841

Mathematics

2 answers:

jasenka [17]2 years ago

8 0

62 should be the answe

WARRIOR [948]2 years ago

3 0

Answer:

7.9, 4.8, 29

Step-by-step explanation:

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Which point lies inside the circle with equation (x-2)^2+(y+3)^2=4

arsen [322]

(x-h)^2+(y-k)^2=r^2

(h,k) is center
r=radius

(x-2)^2+(y-(-3))^2=2^2

center is (2,-3)
radius is 2

A. (2,-5) is on the circle, not inside
B. (2,-4) is inside
C. (0,-3) is on the circle, not inside
D. (2,0) is on the circle, not inside

answer is B

4 0

3 years ago

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:

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