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barxatty [35]
3 years ago
5

A ladder (line segment AC in the diagram) is leaning against a wall. The distance between the foot of the ladder and the wall (B

C) is 7 meters less than the distance between the top of the ladder and the ground (AB). A-Create an equation that models the length of the ladder (l) in terms of x, which is the length in meters of AB. B-If the length of the ladder is 13 meters, use the equation you wrote to find the distance between the ground and the top of the ladder (AB).

Mathematics
1 answer:
Serga [27]3 years ago
8 0

<em>Greetings from Brasil...</em>

a)

Let's just use Pythagoras

L² = X² + (X - 7)²

<h3>L = √(2X² - 14X + 49)</h3><h3 />

b)

If L = 13, then what is the value of X ???

L² = 2X² - 14X + 49

2X² - 14X - 120 = 0

X = 12  or  X = - 5

<em>(The distance cannot be negative, so X = 12)</em>

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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
Can you help me with this.
user100 [1]

I hope this helps you

7 0
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Write a division story with an answer of 1/4 plz ASAP
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Answer: there were 4 people at a party. There was 1 cookie left on the table and all of them wanted it. They divided it between the four of them so they could all have a peice.

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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a line labeled g of x that passes through points negative 1, negative 2 and 0, 2

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Nvm I got it now lol
Usimov [2.4K]

Answer:

nice bro good for you :)

Step-by-step explanation:

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