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navik [9.2K]
3 years ago
6

Please help !! i’ll give 5 stars if help.

Mathematics
1 answer:
Firlakuza [10]3 years ago
6 0

Answer:

true

Step-by-step explanation:

there are two x-intercepts

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Suppose that y varies inversely with x, and y=3 when x=4.
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When y varies inversely as x it can be expressed as follows:

\frac{1}{x}\text{ }\propto\text{y}

If we write the general equation for this

y=\frac{A}{x}

When y=3 and x=4 we can calculate de value of A

\begin{gathered} x=4 \\ y=3 \\ 3=\frac{A}{4} \\ A=4\cdot3 \\ A=12 \end{gathered}

Then the equation that describes an inverse variation between x and y is_

y=\frac{12}{x}

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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-
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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.
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3 years ago
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LenKa [72]

Answer:

4

Step-by-step explanation:

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