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4 years ago

Please help I need answer to this problem.

Mathematics

2 answers:

Svet_ta [14]4 years ago

5 0

Answer:

d 3.38

Step-by-step explanation:

seropon [69]4 years ago

4 0

Are the first numbers ratios?

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D Find mZW. X WK (11x - 29) Z (6x + 5) mZW

Artist 52 [7]

Answer:

m<W = 103°

Step-by-step explanation:

First, find the value of x

m<W + m<Y = 180° (opposite angles in a cyclic quadrilateral are supplementary)

11x - 29 + 6x + 5 = 180 (substitution)

Add like terms

17x - 24 = 180

Add 24 to both sides

17x = 180 + 24

17x = 204

Divide both sides by 17

17x/17 = 204/17

x = 12

Find m<W:

m<W = 11x - 29

Plug in the value of x

m<W = 11(12) - 29 = 132 - 29

m<W = 103°

3 0

3 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:

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