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3 years ago

I needa know how to do this

Mathematics

1 answer:

Elena L [17]3 years ago

5 0

Answer:

So I (3,0) becomes I' (1,3)

H (5,2) H' (1,1)

G (2,4) G' (2,1)

Step-by-step explanation:

So let's start with I, just count how many times you move to the left then how many times you move up.

UR WELCOME BABEEEE~ <3

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2 years ago

Carran is building a fence around his rectangular yard. The fence will cover the front and the 2 sides of the yard. The

dolphi86 [110]

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Answer:

sides: 12 feet
front: 28 feet

Step-by-step explanation:

Let x represent the length of one side of the yard. Then the width of the yard is 2x+4, and the perimeter is ...

P = 2(L +W) . . . . . . . . . . formula for perimeter of a rectangle

80 = 2(x + (2x+4)) . . . . . with values from the problem filled in

40 = 3x +4 . . . . . . . . divide by 2, collect terms

36 = 3x . . . . . . . . . subtract 4

12 = x . . . . . . . . . divide by 3. This is the length of the side fence.

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2 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

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3 years ago

Read 2 more answers

In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$

This gives the infinite many solution.

5 0

3 years ago