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VladimirAG [237]
2 years ago
9

HELP ME FAST I ONLY HAVE 5 MINS LEFT Wanda created the poster shown below:

Mathematics
1 answer:
kirza4 [7]2 years ago
3 0
Your answer is C
Length=8cm and width=9cm
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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
How do you find the length in feet of a circle
Alika [10]

there is no length in a circle

4 0
3 years ago
2. what value of b makes the following equation true
Svetradugi [14.3K]

Answer:

  2. 10

  3. D.  1 for all n

Step-by-step explanation:

2. The applicable rules of exponents are ...

  (a^b)(a^c) = a^(b+c)

  a^b = 1/a^-b

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  \dfrac{4^8}{(4^2)^{-3}}\div 4^4=\dfrac{4^8}{4^{-6}\cdot4^4}=\dfrac{4^8}{4^{-2}}\\\\=4^8\cdot4^2=4^{10}

The value of n is 10.

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3. Using the above rules of exponents, the expression simplifies to ...

  6^(-n+n) = 6^0 = 1

The value is 1 for any n.

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3 years ago
spencer spent 4 hours doing chores over the weekend. If he spent 2/3 for an hour on each chore, how many chores did he do.
Elina [12.6K]
2/3 + 2/3 + 2/3 + 2/3
8/3
2 2/3
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Plz help sovle these one-step equations
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The answer to the question

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