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dimulka [17.4K]

4 years ago

5

Find the coordinates of quadrilateral V' W' X' Y' after a dilation with the scale factor of 2. Original coordinates: V(6, 2), W(

–2, 4), X(–3, –2), Y(3, –5), scale factor of 2

2 answers:

ArbitrLikvidat [17]4 years ago

8 0

V(3,1) w(-1,2) x(-1.5,-1) y(1.5,-2.5)

Elden [556K]4 years ago

6 0

Answer:

V (12,4), W (-4,8), X (-6,-4), Y(6, -10)

Step-by-step explanation:

a dilation with scale of 2 means you would multiply the numbers by 2.

(Only in a reduction would you divide by 2. )

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3 years ago

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Setler [38]

Answer:

a. 62 miles/hour

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Step-by-step explanation:

a. ratio of miles to the number of hours is 434 to 7 can also be written as

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7 0

3 years ago

If 8 = 27 and F = 540, find ). Round to the nearest tenth.

pychu [463]

Answer: F=160

Step-by-step explanation:

if

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4 0

3 years ago

How do I solve x for midsegments

satela [25.4K]

You basically just divide what ever number by 2

7 0

4 years ago

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:

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

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

4 0

4 years ago