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2 years ago

Evaluate 15/r-1 when r=5

Mathematics

2 answers:

DochEvi [55]2 years ago

6 0

Answer: 15/5=3 3-1= 2

The answer would be 2. :P

kkurt [141]2 years ago

4 0

15/r-1 when r=5

15 /(5-1)

15 / 4

3 3/4

15/r-1 when r=5

15 /5 -1

3 -1

it depends on whether the entire term is in the denominator

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Jerry was using matrices to solve the system of three equations. He has shown all his steps. Did he make a mistake if so in what

melomori [17]

There are several ways to solve a system of equation; one of these ways is the use of matrix.

<em>Jerry made a mistake at step 2</em>

From the attachment, the step 1 is represented as:

$\mathbf{\frac 12R_1 \left[\begin{array}{cccc}1&1&1&0\\5&3&-2&-4\\3&2&1&1\end{array}\right] }$

The equation in step 2 is given as:

$\mathbf{R_2 = 5R_1 - R_2}$

This means that:

We subtract the elements of row 2 from the elements of row 1, multiplied by 5

So, we have:

$\mathbf{R_2 = 5\left[\begin{array}{cccc}1&1&1&0\end{array}\right] } - \left[\begin{array}{cccc}5&3&-2&-4\end{array}\right] }$

Expand

$\mathbf{R_2 = \left[\begin{array}{cccc}5&5&5&0\end{array}\right] } - \left[\begin{array}{cccc}5&3&-2&-4\end{array}\right] }$

Subtract corresponding cells

$\mathbf{R_2 = \left[\begin{array}{cccc}0&2&7&4\end{array}\right] }$

So, the new row 2 elements should be:

$\mathbf{ \left[\begin{array}{cccc}0&2&7&4\end{array}\right] }$

However, the row 2 elements of Jerry's steps are:

$\mathbf{R_2 = \left[\begin{array}{cccc}0&-2&-7&-4\end{array}\right] }$

This means that:

Jerry made a mistake; and the mistake is at step 2

Read more about matrix at:

brainly.com/question/21848291

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dalvyx [7]

Answer:

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Step-by-step explanation:

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