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2 years ago

For each tank explain whether or not there is a proportional relationship between the

Mathematics

1 answer:

WARRIOR [948]2 years ago

4 0

Answer:

1x 900y A

1.5xthere is orgian for B but none for a

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Jojo likes to paint She estimates the number of paintings she completes using the function P(W) = w – 2. where w is the number o

KATRIN_1 [288]

Answer:

P[J(y)] = 2/3 * J(y) -2

Step-by-step explanation:

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2 years ago

A truck can travel 12 miles on a gallon of gasoline. How far can the truck travel on 40 gallons of gasoline? Which operation is

deff fn [24]

Answer:

im sorry

Step-by-step explanation:

i dont kno

3 0

2 years ago

What is the area for this?

mote1985 [20]

12.5 (area of triangle) +40 (area of rectangle)=52.5

7 0

1 year ago

Please help and include explanation !!

Sidana [21]

The vertical angles are equal to each other, that is, one of the one-sided corners will be 116°.

The sum of the one-sided angles is 180°, which means: 4x+116 = 180°

4x = 180-116 => 4x = 64°

x = 64/4 = 16°.

The sum of two adjacent angles is 180°. which means, that: 2x-3y = 180-64 = 116°

2x-3y = 116°

-2x+3y = -116

-32+3y = -116

3y = -116-(-32)

3y = -84

y = -84/3 => y = -28.

6 0

2 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