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Umnica [9.8K]
3 years ago
5

9 cm

Mathematics
1 answer:
olga2289 [7]3 years ago
4 0

Step-by-step explanation:

583,200 cubic centimeters

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Natasha places a mirror on the ground 24 ft from the base of an oak
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Answer:

33 ft

Step-by-step explanation:

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x=33

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3 years ago
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Find a formula for the least squares solution of ax=b when the columns of a are orthonormal1 .
neonofarm [45]
Let \mathbf A be a rectangular m\times n matrix with column vectors \mathbf a_1,\ldots,\mathbf a_n, i.e.

\mathbf A=\begin{bmatrix}\mathbf a_1&\cdots&\mathbf a_n\end{bmatrix}

Then we have

\mathbf A^\top=\begin{bmatrix}\mathbf a_1&\cdots&\mathbf a_n\end{bmatrix}^\top

and the product of the two is

\mathbf A^\top\mathbf A=\begin{bmatrix}\mathbf a_1\cdot\mathbf a_1&\mathbf a_1\cdot\mathbf a_2&\cdots&\mathbf a_1\cdot\mathbf a_n\\\mathbf a_2\cdot\mathbf a_1&\mathbf a_2\cdot\mathbf a_2&\cdots&\mathbf a_2\cdot\mathbf a_n\\\vdots&\vdots&\ddots&\vdots\\\mathbf a_n\cdot\mathbf a_1&\mathbf a_n\cdot\mathbf a_2&\cdots&\mathbf a_n\cdot\mathbf a_n\end{bmatrix}

Because the columns of \mathbf A are orthonormal, we have

\mathbf a_i\cdot\mathbf a_j=\begin{cases}1&\text{for }i=j\\0&\text{for }i\neq j\end{cases}

which means \mathbf A^\top\mathbf A reduces to an n\times n matrix with ones along the diagonal and zero everywhere else, i.e.

\mathbf A^\top\mathbf A=\begin{bmatrix}1&0&\cdots&0\\0&1&\cdots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\cdots&1\end{bmatrix}=\mathbf I_n

where \mathbf I denotes the identity matrix. This means the solution to \mathbf{Ax}=\mathbf b is given by

\mathbf A^\top(\mathbf{Ax})=\mathbf A^\top\mathbf b\implies(\underbrace{\mathbf A^\top\mathbf A}_{\mathbf I})\mathbf x=\mathbf A^\top\mathbf b\implies\mathbf x=\mathbf A^\top\mathbf b
6 0
3 years ago
√secA+tanA/√secA-tanA × √cosecA-1/√cosecA+1=1
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Use:\\\\\sec A=\dfrac{1}{\cos A}\\\\\tan A=\dfrac{\sin A}{\cos A}\\\\\csc A=\dfrac{1}{\sin A}\\\\\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}\\---------------------------------\\\\\sec A+\tan A=\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}=\dfrac{1+\sin A}{\cos A}\\\\\sec A-\tan A=\dfrac{1-\sin A}{\cos A}\\\\\csc A-1=\dfrac{1}{\sin A}-\dfrac{\sin A}{\sin A}=\dfrac{1-\sin A}{\sin A}\\\\\csc A+1=\dfrac{1+\sin A}{\sin A}


\dfrac{\sqrt{\sec A+\tan A}}{\sqrt{\sec A-\tan A}}\cdot\dfrac{\sqrt{\cos A-1}}{\sqrt{\cos A+1}}=1\\\\L_s=\sqrt{\dfrac{\sec A+\tan A}{\sec A-\tan A}\cdot\dfrac{\cos A-1}{\cos A+1}}=\sqrt{\dfrac{\frac{1+\sin A}{\cos A}}{\frac{1-\sin A}{\cos A}}\cdot\dfrac{\frac{1-\sin A}{\sin A}}{\frac{1+\sin A}{\sin A}}}\\\\=\sqrt{\dfrac{1+\sin A}{\cos A}\cdot\dfrac{\cos A}{1-\sin A}\cdot\dfrac{1-\sin A}{\sin A}\cdot\dfrac{\sin A}{1+\sin A}}\\\\\text{Everything are simplified}\\\\=\sqrt{1}=1=R_s

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combine like terms

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x=23.5

or

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Irina-Kira [14]

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hope this helps!!!

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