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3 years ago

Which expression is a factor of x2 +2x - 15 A. (x - 3) B. (x + 3) C. (x + 15) D. (x - 5)

Mathematics

1 answer:

valina [46]3 years ago

3 0

Answer:

A. (x - 3)

Step-by-step explanation:

x^2 +2x - 15

Factor

What 2 numbers multiply to -15 and add to 2

5*-3 = -15

5 -3 =2

(x+5)(x-3)

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1. substitution....because it would be easier to just sub one y in for the other then it would be to use the elimination method

2. substitution....because in one of the equations, u already have y...so just sub 6x - 13 in for y in the other equation

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4. substitution....sub in x + 1 for y in the other equation
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the solution to this system is : (-2,-5)

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4 years ago

LINEAR ALGEBRA

kenny6666 [7]

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:

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

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

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

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

The following system of linear equations is obtained:

$-2\cdot \lambda_{1}-4\cdot \lambda_{2}= 1$ (Eq. 3)

$-3\cdot \lambda_{1}-\lambda_{2}= 2$ (Eq. 4)

$-\lambda_{1}-k = 0$ (Eq. 5)

The solution of this system is:

$\lambda_{1} = -\frac{7}{10}$ , $\lambda_{2} = \frac{1}{10}$ , $k = \frac{7}{10}$

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