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2 years ago

Write 0.25x+1y=12 in slope-intercept form.

Mathematics

1 answer:

Pepsi [2]2 years ago

6 0

Answer:

y = -.25x + 12

Step-by-step explanation:

Subtract x on both sides.

y = -.25x + 12

There is your answer. The "1" before the y doesn't need to be written, as a rule in algebra, ones do need to be written and are assumed to be there if there is no number at all.

-.25 is the slope and 12 is the y-intercept.

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Kayle and Olivia both just got their first jobs. After working for two weeks they received pay checks. Kayle received her check

MaRussiya [10]

Answer:

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Step-by-step explanation:

ovlivia makes $8.75 per hour while Kayle makes $8.5 per hour

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2 years ago

Describe the process of creating a linear equation using two points and the point-slope form.

faltersainse [42]

First find the slope of the line = (y2 - y1) / (x2 - x1) where the 2 points are (x1, y1 and (x2, t2).

Then substitute the values of the slope (m) and one of the points into the point-slope formula

y - y1 = m(x - x1)

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3 years ago

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Triss [41]

Answer:

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Step-by-step explanation:

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6 0

3 years ago

Find the inverse of the given matrix, if it exists.Aequals=left bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 0 3

BabaBlast [244]

Answer:

$A^{-1}=\left[ \begin{array}{ccc} \frac{1}{9} & \frac{4}{27} & - \frac{2}{27} \\\\ \frac{8}{9} & \frac{5}{27} & \frac{11}{27} \\\\ - \frac{4}{9} & \frac{2}{27} & - \frac{1}{27} \end{array} \right]$

Step-by-step explanation:

We want to find the inverse of $A=\left[ \begin{array}{ccc} 1 & 0 & -2 \\\\ 4 & 1 & 3 \\\\ -4 & 2 & 3 \end{array} \right]$

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 4&1&3&0&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]$

Subtract row 1 multiplied by 4 from row 2

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]$

Add row 1 multiplied by 4 to row 3

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&2&-5&4&0&1\end{array}\right]$

Subtract row 2 multiplied by 2 from row 3

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&-27&12&-2&1\end{array}\right]$

Divide row 3 by −27

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

Add row 3 multiplied by 2 to row 1

$\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

Subtract row 3 multiplied by 11 from row 2

$\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&0&\frac{8}{9}&\frac{5}{27}&\frac{11}{27} \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

As can be seen, we have obtained the identity matrix to the left. So, we are done.

6 0

3 years ago

What is 3/18 written in simplest form explain how you found your answer

GaryK [48]

1/6 because 3 goes into 3 once, and 3 goes into 18 6 times. So the fraction is 1/6

7 0

2 years ago

Read 2 more answers