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inysia [295]
3 years ago
5

A line is perpendicular to y = - x - 2 and intersects the point (-5, 10)

Mathematics
1 answer:
Lady_Fox [76]3 years ago
6 0

Answer:

y = x + 15

Step-by-step explanation:

If two lines are perpendicular to each other, they have opposite slopes.

The first line is y = - x - 2. Its slope is -1. A line perpendicular to this will have a slope of 1.

Plug this value (1) into your standard point-slope equation of y = mx + b.

y = 1x + b or y = x + b

To find b, we want to plug in a value that we know is on this line: in this case, it is (-5, 10). Plug in the x and y values into the x and y of the standard equation.

10 = (-5) + b

Simplify.

10 = -5 + b

Now, add 5 to both sides to isolate b.

15 = b

Plug this into your standard equation.

y = x + 15

This equation is perpendicular to your given equation (y = -x - 2) and contains point (-5, 10)

Hope this helps!

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Step-by-step explanation:

If we have the following expression: 15 - (37 + 8) : 3.

We will first divide the value we get in parentheses and then subtract it.

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I need help with a math equation, it’s due by 3:00.
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Step-by-step explanation:

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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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