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3 years ago

Select the graph of the solution. Click until the correct graph appears. |x| + 1 ≤ 2

Mathematics

2 answers:

Nikolay [14]3 years ago

3 0

|x| + 1 < 2

minus one on both sides

|x| < 1

This represents a closed circle around one going positive one going left. However based on the images choose the one with colored in circles going outwards from the -1 and 1.

Hope this helps! If you think I'm worthy, Brainliest would be great! :D

zmey [24]3 years ago

3 0

|x| + 1 < 2

That is the answer

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Answer:

The value of the constant $k$ so that $\vec u_{3}$ is a linear combination of $\vec u_{1}$ and $\vec u_{2}$ is $\frac{7}{10}$ .

Step-by-step explanation:

Let be $\vec u_{1} = [2,3,1]$ , $\vec u_{2} = [4,1,0]$ and $\vec u_{3} = [1, 2,k]$ , $\vec u_{3}$ is a linear combination of $\vec u_{1}$ and $\vec u_{3}$ if and only if:

$\alpha_{1} \cdot \vec u_{1} + \alpha_{2} \cdot \vec u_{2} +\alpha_{3}\cdot \vec u_{3} = \vec O$ (Eq. 1)

Where:

$\alpha_{1}$ , $\alpha_{2}$ , $\alpha_{3}$ - Scalar coefficients of linear combination, dimensionless.

By dividing each term by $\alpha_{3}$ :

$\lambda_{1}\cdot \vec u_{1} + \lambda_{2}\cdot \vec u_{3} = -\vec u_{3}$

$\vec u_{3}=-\lambda_{1}\cdot \vec u_{1}-\lambda_{2}\cdot \vec u_{2}$ (Eq. 2)

$\vec O$ - Zero vector, dimensionless.

And all vectors are linearly independent, meaning that at least one coefficient must be different from zero. Now we expand (Eq. 2) by direct substitution and simplify the resulting expression:

$[1,2,k] = -\lambda_{1}\cdot [2,3,1]-\lambda_{2}\cdot [4,1,0]$