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yanalaym [24]
3 years ago
5

8(x+3)=2xsolve for x​

Mathematics
2 answers:
kvv77 [185]3 years ago
6 0

Answer: x = -4

Step-by-step explanation: The 1st graders at City Elementary were asked whether they like dogs or cats best. The results are shown in the table.

Relative Frequency Table by Row

A 4-column table with 2 rows. Column 1 has entries boys, girls. Column 2 is labeled Dogs with entries 42.1 percent, 79.8 percent. Column 3 is labeled Cats with entries 57.9 percent, 20.2 percent. Column 4 is labeled Total with entries 100 percent, 100 percent.

What conclusion can you draw about the relative frequency of the results?

A girl in this group is most likely to prefer cats.

A girl in this group is most likely to prefer dogs.

A boy in this group is most likely to prefer dogs.

There is no association between the variables.

WINSTONCH [101]3 years ago
3 0

Answer:

x = -4

Step-by-step explanation:

8 (x+3) = 2x

8x + 8*3 = 2x

8x + 24 = 2x

6x = -24

x = -4

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Answer:

 

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Step-by-step explanation:

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Let [v1,v2,v3] be a set of nonzero vectors in r^m such that the (transpose of vi)*vj = 0 when i is not equal to j. show that the
babunello [35]
Let \mathbf V be the m\times3 matrix whose columns are \mathbf v_1,\mathbf v_2,\mathbf v_3, and let \mathbf c be the vector whose components are the constants c_1,c_2,c_3. Now consider the matrix equation

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Multiplying both sides by \mathbf V^\top, we have

\mathbf V^\top(\mathbf V\mathbf c)=(\mathbf V^\top\mathbf V)\mathbf c=\mathbf 0

More explicitly, we're writing

\mathbf V=\begin{bmatrix}\mathbf v_1&\mathbf v_2&\mathbf v_3\end{bmatrix}

Multiply both sides by \mathbf V^\top and the left hand side can be written as

\mathbf V^\top\mathbf V=\begin{bmatrix}{\mathbf v_1}^\top\\{\mathbf v_2}^\top\\{\mathbf v_3}^\top\end{bmatrix}\begin{bmatrix}\mathbf v_1&\mathbf v_2&\mathbf v_3\end{bmatrix}=\begin{bmatrix}{\mathbf v_1}^\top\mathbf v_1&{\mathbf v_1}^\top\mathbf v_2&{\mathbf v_1}^\top\mathbf v_3\\{\mathbf v_2}^\top\mathbf v_1&{\mathbf v_2}^\top\mathbf v_2&{\mathbf v_2}^\top\mathbf v_3\\{\mathbf v_3}^\top\mathbf v_1&{\mathbf v_3}^\top\mathbf v_2&{\mathbf v_3}^\top\mathbf v_3\end{bmatrix}

We're told that {\mathbf v_i}^\top\mathbf v_j=0 whenever i\neq j, so we're left with

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Each of \mathbf v_1,\mathbf v_2,\mathbf v_3 are nonzero, which means their norms are nonzero, which necessarily implies that \mathbf c=0, and so the vectors \mathbf v_1,\mathbf v_2,\mathbf v_3 must necessarily be linearly independent.
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Answer:

Attachment!

Step-by-step explanation:

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Answer:

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