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AfilCa [17]
3 years ago
6

16055 divided by16 what is the remainder

Mathematics
2 answers:
nasty-shy [4]3 years ago
8 0
The remainder is 7
Hope this helps
puteri [66]3 years ago
6 0
103 r7 Hope this helped!
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Yes bro hope it helped give me a thanks
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You are having a conference call with the CEO of a paper company. You have interpreted the number of trees cut down versus profi
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I need so much help with these could you help me?
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Help please answer questions below
djverab [1.8K]
3/8 is0.375 so it would stay as 2 and 5/6 is 0.8333 so it rounds up to 9
So now do 2 times 9 = 18
4 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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