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Rama09 [41]
2 years ago
11

The standard ratio of a photo's width to it's length is 2/3. what is the length of a photo that has a width of 14 inches?

Mathematics
2 answers:
ankoles [38]2 years ago
5 0
L: w

Equation

2/3 = 14/ x

Cross multiply

2xX = 14x3

2x = 42

Eliminate

2x/2 = 42/2

X = 21
AysviL [449]2 years ago
5 0

Answer:

The answer is D.21.

Step-by-step explanation:

You have to divide 14 by 2/3 and you get 21.

PLZ mark me Brainliest!

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Determine if the columns of the matrix form a linearly independent set. Justify your answer. [Start 3 By 4 Matrix 1st Row 1st Co
Volgvan

Answer:

Linearly Dependent for not all scalars are null.

Step-by-step explanation:

Hi there!

1)When we have vectors like v_{1},v_{2},v_{3}, ... we call them linearly dependent if we have scalars a_{1},a_{2},a_{3},... as scalar coefficients of those vectors, and not all are null and their sum is equal to zero.

a_{1}\vec{v_{1}}+a_{2}\vec{v_{2}}+a_{3}\vec{v_{3}}+...a_{m}\vec{v_{m}}=0  

When all scalar coefficients are equal to zero, we can call them linearly independent

2)  Now let's examine the Matrix given:

\begin{bmatrix}1 &-2  &2  &3 \\ -2 & 4 & -4 &3 \\ 0&1  &-1  & 4\end{bmatrix}

So each column of this Matrix is a vector. So we can write them as:

\vec{v_{1}}=\left \langle 1,-2,1 \right \rangle,\vec{v_{2}}=\left \langle -2,4,-1 \right \rangle,\vec{v_{3}}=\left \langle 2,-4,4 \right \rangle\vec{v_{4}}=\left \langle 3,3,4 \right \rangle Or

Now let's rewrite it as a system of equations:

a_{1}\begin{bmatrix}1\\ -2\\ 0\end{bmatrix}+a_{2}\begin{bmatrix}-2\\ 4\\ 1\end{bmatrix}+a_{3}\begin{bmatrix}2\\ -4\\ -1\end{bmatrix}+a_{4}\begin{bmatrix}3\\ 3\\ 4\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}

2.1) Since we want to try whether they are linearly independent, or dependent we'll rewrite as a Linear system so that we can find their scalar coefficients, whether all or not all are null.

Using the Gaussian Elimination Method, augmenting the matrix, then proceeding the calculations, we can see that not all scalars are equal to zero. Then it is Linearly Dependent.

 \left ( \left.\begin{matrix}1 &-2  &2  &3 \\ -2 &4  &-4  &3 \\ 0 & 1 &-1  &4 \\ \end{matrix}\right|\begin{matrix}0\\ 0\\ 0\end{matrix} \right )R_{1}\times2 +R_{2}\rightarrow R_{2}\left ( \left.\begin{matrix}1 &-2  &2  &3 \\ 0 &0 &9  &0\\ 0 & 1 &-1  &4 \\ \end{matrix}\right|\begin{matrix}0\\ 0\\ 0\end{matrix} \right )\ R_{2}\Leftrightarrow  R_{3}\left ( \left.\begin{matrix}1 &-2  &2  &3 \\ 0 &1  &-1  &4 \\ 0 &0 &9  &0 \\ \end{matrix}\right|\begin{matrix}0\\ 0\\ 0\end{matrix} \right )\left\{\begin{matrix}1a_{1} &-2a_{2}  &+2a_{3}  &+3a_{4}  &=0 \\  &1a_{2}  &-1a_{3} &+4a_{4}  &=0 \\  &  &  &9a_{4}  &=0 \end{matrix}\right.\Rightarrow a_{1}=0, a_{2}=a_{3},a_{4}=0

S=\begin{bmatrix}0\\ a_{3}\\ a_{3}\\ 0\end{bmatrix}

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