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3 years ago

Complete the equation for the standard form of the line that has an x-intercept of 3 and a y-intercept of 5.

Mathematics

2 answers:

atroni [7]3 years ago

6 0

5x+3y=15

Enjoy your day my friend!!

Ugo [173]3 years ago

3 0

Answer:

5x+3y=15

Step-by-step explanation:

We are given two coordinates (3,0) (0,5)

Start by finding slope

y2-y1/x2-x1= 5-0/0-3= -5/3 this is your slope m.

y=mx+c

y= -5/3x+c (but remember standard line is ax+by=c) so we have to switch sides

5/3x + y=5 (remember standard form has no fraction so multiply by 3 on all sides)

5x+3y=15

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3 years ago

Please Help I don’t get it at all N/2+5=3

eduard

N/2+5=3

move +5 to the other side

sign changes from +5 to -5

N/2+5-5= 3-5

N/2= -2

Mutiply both sides by 2/1

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3 years ago

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BigorU [14]

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3 years ago

For the following linear system, put the augmented coefficient matrix into reduced row-echelon form.

Anni [7]

Answer:

The reduced row-echelon form of the linear system is $\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right]$

Step-by-step explanation:

We will solve the original system of linear equations by performing a sequence of the following elementary row operations on the augmented matrix:

Interchange two rows
Multiply one row by a nonzero number
Add a multiple of one row to a different row

To find the reduced row-echelon form of this augmented matrix

$\left[\begin{array}{cccc}2&3&-1&14\\1&2&1&4\\5&9&2&7\end{array}\right]$

You need to follow these steps:

Divide row 1 by 2 $\left(R_1=\frac{R_1}{2}\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\1&2&1&4\\5&9&2&7\end{array}\right]$

Subtract row 1 from row 2 $\left(R_2=R_2-R_1\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\5&9&2&7\end{array}\right]$

Subtract row 1 multiplied by 5 from row 3 $\left(R_3=R_3-\left(5\right)R_1\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]$

Subtract row 2 multiplied by 3 from row 1 $\left(R_1=R_1-\left(3\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]$

Subtract row 2 multiplied by 3 from row 3 $\left(R_3=R_3-\left(3\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&0&0&-19\end{array}\right]$

Multiply row 2 by 2 $\left(R_2=\left(2\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&-19\end{array}\right]$

Divide row 3 by −19 $\left(R_3=\frac{R_3}{-19}\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&1\end{array}\right]$

Subtract row 3 multiplied by 16 from row 1 $\left(R_1=R_1-\left(16\right)R_3\right)$

$\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&-6\\0&0&0&1\end{array}\right]$

Add row 3 multiplied by 6 to row 2 $\left(R_2=R_2+\left(6\right)R_3\right)$

$\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right]$

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densk [106]

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3 years ago

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