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yaroslaw [1]
3 years ago
5

The equation for the line g can be written as x-y=9. Line h is parallel to line g and passes through (4,5). What is the equation

of line h?
Write the equation in slope-intercept form. Write the numbers in the equation as proper fractions, improper fractions, or integers.

Please help me!!!
Mathematics
1 answer:
Gnoma [55]3 years ago
8 0

Answer:

y = x+1

Step-by-step explanation

The equation in slope intercept form is expressed as y = MX+c

m a the slope

Get the slope of the line

Given the line x-y =9

-y = -x+9

y = x-9

Slope of the line is 1

Since it is parallel to the given line, the slope o the required line is also 1

Substitute m = 1 adnd (4,5) into the point slope form

y-y0 =m(x-x0)

y - 5 = 1(x-4)

y-5 = x-4

y= x-4+5

y = x+1

Hence the required equation is y = x+1

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Read 2 more answers
For waht values of x do the vectors -1,0,-1), (2,1,2), (1,1, x) form a basis for R3?
DaniilM [7]
<h2>Answer:</h2>

The values of x for which the given vectors are basis for R³ is:

                        x\neq 1

<h2>Step-by-step explanation:</h2>

We know that for a set of vectors are linearly independent if the matrix formed by these set of vectors is non-singular i.e. the determinant of the matrix formed by these vectors is non-zero.

We are given three vectors as:

(-1,0,-1), (2,1,2), (1,1, x)

The matrix formed by these vectors is:

\left[\begin{array}{ccc}-1&2&1\\0&1&1\\-1&2&x\end{array}\right]

Now, the determinant of this matrix is:

\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-1(x-2)-2(1)+1\\\\\\\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-x+2-2+1\\\\\\\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-x+1

Hence,

-x+1\neq 0\\\\\\i.e.\\\\\\x\neq 1

4 0
3 years ago
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