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3 years ago

Independent Practice

Mathematics

1 answer:

Sedbober [7]3 years ago

5 0

Answer: THE ANSWER IS D. (-27, -7, -2, 8, 48)

Step-by-step explanation:

Graph.

y = 5 x − 2

{ − 5 , − 1 , 0 , 2 , 10 }

The chosen topic is not meant for use with this type of problem. Try the examples below.

1 , 2 , 2 , 2 , 3 , 4 , 4 , 5 , 8 , 9 , 10

11 , 14 , 14 , 17 , 17 , 17 , 41 , 44 , 47 , 71 , 74 , 77

12 , 15 , 45 , 65 , 78

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Irene paid $40 more of her cell phone bill in July then in June and August she paid $10 less than her June bill if she paid a to

MAXImum [283]

She paid 50.00 for her august bill

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Kedar eaens a monthly salary of $2,200 plus a 3.75% commission on the amount of his sales at a men's clothing store . What would

NemiM [27]

Okay. So, Kedar earns a monthly salary of $2,200 per month. That stays the same. He earns a 3.75% monthly commission on how much he sells. To find out how much his commission would be if he sold $4,500 in clothing, we would have to multiply 4,500 by 3.75% (0.0375). When you multiply both numbers, you get $168.75 in commission sales. Now, you must add that plus 2,200. When you add both of those numbers, you get 2,368.75. Kedar would earn $2,368.75 this month if he sold $4,500 worth of clothes.

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4 years ago

25 POINTS PLZ PLZ HELP ME! I NEED IT

Dimas [21]

Answer:

f (x) is translated 5/3 units left

Step-by-step explanation:

I think thats Right

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4 years ago

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A dermatology patient is treated with ultraviolet light having a wavelength of 375 nm. What is the wavelength expressed in centi

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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

$\mathbf A^\top(\mathbf{Ax})=\mathbf A^\top\mathbf b\implies(\underbrace{\mathbf A^\top\mathbf A}_{\mathbf I})\mathbf x=\mathbf A^\top\mathbf b\implies\mathbf x=\mathbf A^\top\mathbf b$

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3 years ago