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2 years ago

Solve for the unknown variable "x" in the following linea equation: 2 ( 2x + 20) = 80 X =

Mathematics

1 answer:

Stella [2.4K]2 years ago

4 0

Step-by-step explanation:

2(2+20)=80

4+40=80 (Distribute)

4+40-40=80-40(Subtract from both sides of the equation)

4=40(Simplify)

4/4=404/4(Divide both sides of the equation by the same term to get x=10)

=10

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Convert 7.7km into miles if 1.6 km = 1 mile

Lemur [1.5K]

This is a simple equation.

The equation is y = 1.6x with y being the miles and x being the kilometers.

Lets plug the kilometers in.

Y=1.6(7.7)

1.6 * 7.7 = 12.32

12.32 miles is equal to 7.7 kilometers

HOWEVER

There are about .6 miles in a kilometer, so if you typed this wrong I don't know.

It is still the same concept.

Y = .6x

Y = .6(7.7)

.6 * 7.7 = 4.8

In this case there are 4.8 miles in 7.7 kilometers

3 0

2 years ago

98 POINTS!!!!! PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Maslowich

Hello!

I believe the answer is C.

I hope it helps!

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3 years ago

Read 2 more answers

The sum of matrices A and B is C.

GuDViN [60]

Answer:

Step-by-step explanation: -3

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2 years ago

Answer correctly please , will mark brainliest answer !!!!

Afina-wow [57]

Answer:

x-2

Step-by-step explanation:

(7x^2-14x)/7x

Distribute the 7

x-2

7 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

2 years ago