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3 years ago

PLEASE HELP ME...

Mathematics

2 answers:

adell [148]3 years ago

4 0

Since the reflected figure is moved to the third quardrant, the reflection should be across the x-axis.

Also, if we compare the width and height of the orginial triangle with that of the reflected one, it is clear that it is dilated with a scale factor of 0.25.

erastovalidia [21]3 years ago

4 0

Answer:

Triangle ABC can be mapped onto triangle DEF with a reflection across the x axis followed by a dilation centered at the origin with a scale factor of 0.25 .

Step-by-step explanation:

Given : Graph of two triangle ABC and DEF with coordinates .

To find : Triangle ABC can be mapped onto triangle DEF with a reflection across the______ followed by a dilation centered at the origin with a scale factor of ______.

Solution : We can see from given graph triangle triangle ABC is reflected across x axis.

Scale factor: 0.25

Therefore, Triangle ABC can be mapped onto triangle DEF with a reflection across the x axis followed by a dilation centered at the origin with a scale factor of 0.25 .

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Find the values of λ for which the determinant is zero.

elena55 [62]

Answer:

Determinant are special number that can only be defined for square matrices.

Step-by-step explanation:

Determinant are particularly important for analysis. The inverse of a matrix exist, if the determinant is not equal to zero.

How to find determinant

For a 2×2 matrix

$det ( \left[\begin{array}{cc}x&y\\a&z\end{array}\right] ) = xz-ay$

For a 3×3 matrix

we first decompose it to 2×2

$det (\left[\begin{array}{ccc}k&l&m\\o&p&q\\r&s&t\end{array}\right] )\\\\= k*det(\left[\begin{array}{cc}p&q\\s&t\end{array}\right] ) - l*det(\left[\begin{array}{cc}o&q\\r&t\end{array}\right] ) + m*det(\left[\begin{array}{cc}o&p\\r&s\end{array}\right] ) \\\\=k(pt-sq) - l(ot-rq) + m(os-rp)$

Example

Find the values of λ for which the determinant is zero

$\left[\begin{array}{ccc}s&-1&0\\-1&s&-1\\0&-1&1\end{array}\right]$

$det(\left[\begin{array}{ccc}s&-1&0\\-1&s&-1\\0&-1&1\end{array}\right])\\\\= s*det(\left[\begin{array}{cc}s&-1\\-1&1\end{array}\right] ) - (-1)*det(\left[\begin{array}{cc}-1&-1\\0&1\end{array}\right] ) + 0*det(\left[\begin{array}{cc}-1&s\\0&-1\end{array}\right] )\\\\= s(s(1)-(-1*-1)) - (-1)(-1*1 - (-1*0)) + 0\\= s(s - 1)) + 1(-1 + 0) \\=s^{2} -s-1$

Equating the determinant to zero

$s^{2} -s-1 =0\\$

s = $\frac{1}{2}$ * (1 ±5 )

s = 1.61 or -0.61

5 0

3 years ago

Consider a set of one-dimensional points: 6, 12, 18, 24, 30, 42, 48.

Vinvika [58]

Answer:

1) With initial centroids 10-40, final clusters are

first cluster (6,12,18,24,30)

second cluster (42,48)

And the Sum of Squared Errors (SSE) for the clustering result is 378

2) With initial centroids 10-20, final clusters are

first cluster (6,12,18,24)

second cluster (30,42,48)

And the Sum of Squared Errors (SSE) for the clustering result is 348

Step-by-step explanation:

K-means works as follows

the points will be clustered according to their distance to the centroids.
Then centroids are updated as the cluster means.
This process continues until clusters doesn't change anymore

1) initial centroids 10-40

first cluster (6,12,18,24) mean:15

second cluster (30,42,48) mean:40

new centroids 15-40

first cluster (6,12,18,24,30) mean:18

second cluster (42,48) mean:45

final centroids 18-45

first cluster (6,12,18,24,30) mean:18

second cluster (42,48) mean:45

Sum of Squared errors = $(18-6)^{2}$ + $(18-12)^{2}$ + $(18-18)^{2}$ + $(18-24)^{2}$ + $(18-30)^{2}$ + $(45-42)^{2}$ + $(45-48)^{2}$ =378

2)initial centroids 10-20

first cluster (6,12) mean:9

second cluster (18,24,30,42,48) mean:32.4

new centroids 9-32.4

first cluster (6,12,18) mean:12

second cluster (24,30,42,48) mean:36

new centroids 12-36

first cluster (6,12,18,24) mean:15

second cluster (30,42,48) mean:40

final centroids 15-40

first cluster (6,12,18,24) mean:15

second cluster (30,42,48) mean:40

Sum of Squared errors = $(15-6)^{2}$ + $(15-12)^{2}$ + $(15-18)^{2}$ + $(15-24)^{2}$ + $(40-30)^{2}$ + $(40-42)^{2}$ + $(40-48)^{2}$ =348

6 0

3 years ago

Is 6 a factor of 81 yes or no

topjm [15]

To test whether it's a factor:

-- Divide the big number by it.     81 / 6 = 13.5 .

-- If the quotient is a whole number,
   then the small number is a factor
   of the big number. 13.5 is not a whole number.

-- If the quotient is not a whole number,
   then the small number is NOT a factor
   of the big number. 6 is NOT a factor of 81 .

8 0

3 years ago

Read 2 more answers

Plz help i don't understand

gayaneshka [121]

How can i help you?
I would love to help you

4 0

3 years ago

1. A room is in the shape of a rectangle. The width of the room is 2 feet longer than two thirds the length. If the room is 33 f

morpeh [17]

2/3 of 33 is 22
so 22 +2 = 24
so 24x33=792
so the area is 792

7 0

4 years ago