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Snowcat [4.5K]
2 years ago
12

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

step.

Mathematics
2 answers:
grandymaker [24]2 years ago
5 0

Answer:

Step 3

Step-by-step explanation:

melomori [17]2 years ago
4 0

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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Answer:

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Step-by-step explanation:

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Answer and Step-by-step explanation:

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Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species
kotegsom [21]

Answer:

P(t) = \frac{160000e^{1.36t}}{2000 + 80(e^{1.36t} - 1)}

Step-by-step explanation:

The logistic equation is the following one:

P(t) = \frac{KP(0)e^{rt}}{K + P(0)(e^{rt} - 1)}

In which P(t) is the size of the population after t years, K is the carrying capacity of the population, r is the decimal growth rate of the population and P(0) is the initial population of the lake.

In this problem, we have that:

Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 2,000. This means that P(0) = 80, K = 2000.

The number of fish tripled in the first year. This means that P(1) = 3P(0) = 3(80) = 240.

Using the equation for P(1), that is, P(t) when t = 1, we find the value of r.

P(t) = \frac{KP(0)e^{rt}}{K + P(0)(e^{rt} - 1)}

240 = \frac{2000*80e^{r}}{2000 + 80(e^{r} - 1)}

280*(2000 + 80(e^{r} - 1)) = 160000e^{r}

280*(2000 + 80e^{r} - 80) = 160000e^{r}

280*(1920 + 80e^{r}) = 160000e^{r}

537600 + 22400e^{r} = 160000e^{r}

137600e^{r} = 537600

e^{r} = \frac{537600}{137600}

e^{r} = 3.91

Applying ln to both sides.

\ln{e^{r}} = \ln{3.91}

r = 1.36

This means that the expression for the size of the population after t years is:

P(t) = \frac{160000e^{1.36t}}{2000 + 80(e^{1.36t} - 1)}

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