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4 years ago

Given= 3a + 5 = 17

Mathematics

2 answers:

choli [55]4 years ago

5 0

Answer:

a=4

Step-by-step explanation:

you subtract the 5 from both sides and then you divide the 3 into the three and then you have 4

IRISSAK [1]4 years ago

3 0

Answer: a=4

Step-by-step explanation: sorry I don’t have enough time to type out the explanation :(

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3 years ago

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3 years ago

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.

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

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

3 0

3 years ago

Complete the input-output table for linear function y=3x

Veseljchak [2.6K]

The input and output table for the linear function y = 3x is (-2, -6), (-1, -3), (0, 0), (1, 3) and (2, 6).

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables and numbers.

Given the linear function y = 3x:

When x = -2; y = 3(-2) = -6

When x = -1; y = 3(-1) = -3

When x = 0; y = 3(0) = 0

When x = 1; y = 3(1) = 3

When x = 2; y = 3(2) = 6

The input and output table for the linear function y = 3x is (-2, -6), (-1, -3), (0, 0), (1, 3) and (2, 6).

Find out more on equation at: brainly.com/question/2972832

#SPJ1

4 0

2 years ago

5 1/3 added to four times a number is 38 14/15. what is the number? please show work

Lesechka [4]

38 14/15 - 5 1/3 = 33 3/5 or 33.6
33.6/4= 8.4
8.4= 8 2/5

the answer is 8 2/5

6 0

3 years ago

Given= 3a + 5 = 17 ​

Given= 3a + 5 = 17