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ch4aika [34]
3 years ago
13

What is the middle number when the numbers {2.3, –0.1, 1.9, –4, –3.2} are ordered from least to greatest?

Mathematics
2 answers:
Margaret [11]3 years ago
5 0
First, arrange it from least to greatest:

-4, -3.2, <u>-0.1</u>, 1.9, 2.3

So the answer is -0.1.
Lerok [7]3 years ago
3 0

Answer:

4, -3.2, -0.1, 1.9, 2.3Step-by-step explanation:

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Find the quotients. 25÷34 94÷-34 -57÷-13 -53÷16
Assoli18 [71]

Answer:

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spin [16.1K]

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AnnyKZ [126]

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7 0
3 years ago
Determine whether the set of vectors <img src="https://tex.z-dn.net/?f=%20v_%7B1%3D%283%2C2%2C1%29%2C%20v_%7B2%7D%20%3D%28-1%2C-
Korolek [52]
Since each vector is a member of \mathbb R^3, the vectors will span \mathbb R^3 if they form a basis for \mathbb R^3, which requires that they be linearly independent of one another.

To show this, you have to establish that the only linear combination of the three vectors c_1\mathbf v_1+c_2\mathbf v_2+c_3\mathbf v_3 that gives the zero vector \mathbf0 occurs for scalars c_1=c_2=c_3=0.

c_1\begin{bmatrix}3\\2\\1\end{bmatrix}+c_2\begin{bmatrix}-1\\-2\\-4\end{bmatrix}+c_3\begin{bmatrix}1\\1\\-1\end{bmatrix}=(0,0,0)\iff\begin{bmatrix}3&-1&1\\2&-2&1\\1&-4&-1\end{bmatrix}\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}

Solving this, you'll find that c_1=c_2=c_3=0, so the vectors are indeed linearly independent, thus forming a basis for \mathbb R^3 and therefore they must span \mathbb R^3.
4 0
3 years ago
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