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3 years ago

I need help, can anyone help?

Mathematics

1 answer:

Andreas93 [3]3 years ago

5 0

Answer:

(1) ∆DBE =~ ∆CEA by SAS Theorem

(2) ∆ABD =~ ∆ACD by SSS Theorem

Step-by-step explanation:

Hope you got it.

You might be interested in

-|3n| - 2 = 4 a. -8 b. -2 c. 3 d. -3

Ber [7]

Answer:

-2

Step-by-step explanation:

-|3n| - 2 = 4

|3n| = -6

3n = -6 or 3n = -(-6)

n = -2 n = 2

Since 2 is not part of the options, -2 is the only correct answer

6 0

2 years ago

Help. i need itttttt

ElenaW [278]

Answer:

In a parallelogram, opposite sides are the same length. If PR = ST, then the same would happen for the other sides.

Step-by-step explanation:

8 0

3 years ago

The domain of (fg)(x) consists of the numbers x that are in the domains of both f and g.

Dovator [93]

The statement "The domain of (fg)(x) consists of the numbers x that are in the domains of both f and g" is FALSE.

Domain is the values of x in the function represented by y=f(x), for which y exists.

THe given statement is "The domain of (fg)(x) consists of the numbers x that are in the domains of both f and g".

Now we assume the $g(x)=x+2$ and $f(x)=\frac{1}{x-6}$

So here since g(x) is a polynomial function so it exists for all real x.

$f(x)=\frac{1}{x-6}$ does not exists when $x=6$ , so the domain of f(x) is given by all real x except 6.

Now,

$(fg)(x)=f(g(x))=f(x+2)=\frac{1}{(x+2)-6}=\frac{1}{x-4}$

So now (fg)(x) does not exists when x=4, the domain of (fg)(x) consists of all real value of x except 4.

But domain of both f(x) and g(x) consists of the value x=4.

Hence the statement is not TRUE universarily.

Thus the given statement about the composition of function is FALSE.

Learn more about Domain here -

brainly.com/question/2264373

#SPJ10

3 0

1 year 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

Please help me with this math

Lynna [10]

Number 3: 176 squared cm

7 0

2 years ago