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icang [17]
3 years ago
12

Please help I already did the hint part but I’m not sure how to get the width or I don’t even know

Mathematics
2 answers:
Novosadov [1.4K]3 years ago
6 0

Answer:

1 footsie

Step-by-step explanation:

the first thing ur gonna wanna do is set them equal:

4w^2 + 70w = 74

now in order to solve it you are goonna put it all onto one side:

4w^2+ 70w − 74 = 0

Now, you are gonna simplify this:

2w² + 35w − 37 = 0

(w − 1) (2w + 37) = 0

Set each factor to 0 and solve:

w − 1 = 0

w = 1

2w + 37 = 0

w = -18.5

Since w must be positive, w = 1.  so the width must be 1 foot

Julli [10]3 years ago
5 0

Answer:

1 foot

Step-by-step explanation:

Set equal:

4w² + 70w = 74

Move to one side:

4w² + 70w − 74 = 0

Simplify:

2w² + 35w − 37 = 0

Factor.  Using the AC method, ac = 2×-37 = -74.  Factors of -74 that add up to 35 are 37 and -2.  Dividing by a, the factors reduce to 37/2 and -1/1.

(w − 1) (2w + 37) = 0

Set each factor to 0 and solve:

w − 1 = 0

w = 1

2w + 37 = 0

w = -18.5

Since w must be positive, w = 1.  The width of the wood border is 1 foot.

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Step-by-step explanation:

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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
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