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Alexxandr [17]
2 years ago
13

Triangle JKL is dilated with the origin as the center of dilation to create J’K’L’.

Mathematics
2 answers:
maria [59]2 years ago
7 0

It should be D , it’s the answer choices with the fraction 7/4

RideAnS [48]2 years ago
3 0

Answer:yo dude sorry for making you wait im taking the bench mark rn and i needed the answer but i looked at my notes and im 23.142% sure its the 3rd one

Step-by-step explanation:

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Which is equivalent to 8.03 x 10-8?
Veronika [31]

Answer:

16.06

20characterlimitlmasdfdgfds

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3 years ago
BRAINLIEST!!! HELP!!<br> 3. Rotate the triangle counterclockwise about point D.
Ksivusya [100]
The answer is the first one
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3 years ago
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Liam works at a zoo. He was looking at some data showing the masses of their 5 African elephants. The mean mass of the elephants
hram777 [196]

Answer:

Leo has b boxes of pencils. each box contains 6 pencils. he has a total of 42 pencils. the equation that represents this situation the value of b that makes the equation true the first one is b+6=42,6b=42,b=42+6,or 42b=6 the second one are 7,836 48

Step-by-step explanation:

/

4 0
2 years ago
Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In e
Sedaia [141]

The probability that, at the tip of the fourth round, each of the players has four coins is 5/192.

Given that game consists of 4 rounds and every round, four balls are placed in an urn one green, one red, and two white.

It amounts to filling in an exceedingly 4×4 matrix. Columns C₁-C₄ are random draws each round; row of every player.

Also, let \%R_{A} be the quantity of nonzero elements in R_{A}.

Let C_{1}=\left(\begin{array}{l}1\\ -1\\ 0\\ 0\end{array}\right).

Parity demands that \%R_{A} and\%R_{B} must equal 2 or 4.

Case 1: \%R_{A}=4 and \%R_B=4. There are \left(\begin{array}{l}3\\ 2\end{array}\right)=3 ways to put 2-1's in R_A, so there are 3 ways.

Case 2: \%R_{A}=2 and \%R_B=4. There are 3 ways to position the -1 in R_A, 2 ways to put the remaining -1 in R_B (just don't put it under the -1 on top of it!), and a pair of ways for one among the opposite two players to draw the green ball. (We know it's green because Bernardo drew the red one.) we are able to just double to hide the case of \%R_{A}=4,\%R_{B}=2 for a complete of 24 ways.

Case 3: \%R_A=\%R_B=2. There are 3 ways to put the -1 in R_{A}. Now, there are two cases on what happens next.

  • The 1 in R_B goes directly under the -1 inR_A. There's obviously 1 way for that to happen. Then, there are 2 ways to permute the 2 pairs of 1,-1 in R_C andR_D. (Either the 1 comes first inR_C or the 1 comes first in R_D.)
  • The 1 in R_B doesn't go directly under the -1 in R_A. There are 2 ways to put the 1, and a couple of ways to try and do the identical permutation as within the above case.

Hence, there are 3(2+2×2)=18 ways for this case. There's a grand total of 45 ways for this to happen, together with 12³ total cases. The probability we're soliciting for is thus 45/(12³)=5/192

Hence, at the top of the fourth round, each of the players has four coins probability is 5/192.

Learn more about probability and combination is brainly.com/question/3435109

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3 0
2 years ago
Read 2 more answers
Determine whether the statements are True or False. Justify your answer with an explanation.
frez [133]

Problem 1

<h3>Answer: False</h3>

---------------------------------

Explanation:

The notation (f o g)(x) means f( g(x) ). Here g(x) is the inner function.

So,

f(x) = x+1

f( g(x) ) = g(x) + 1 .... replace every x with g(x)

f( g(x) ) = 6x+1 ... plug in g(x) = 6x

(f o g)(x) = 6x+1

Now let's flip things around

g(x) = 6x

g( f(x) ) = 6*( f(x) ) .... replace every x with f(x)

g( f(x) ) = 6(x+1) .... plug in f(x) = x+1

g( f(x) ) = 6x+6

(g o f)(x) = 6x+6

This shows that (f o g)(x) = (g o f)(x)  is a false equation for the given f(x) and g(x) functions.

===============================================

Problem 2

<h3>Answer: True</h3>

---------------------------------

Explanation:

Let's say that g(x) produced a number that wasn't in the domain of f(x). This would mean that f( g(x) ) would be undefined.

For example, let

f(x) = 1/(x+2)

g(x) = -2

The g(x) function will always produce the output -2 regardless of what the input x is. Feeding that -2 output into f(x) leads to 1/(x+2) = 1/(-2+2) = 1/0 which is undefined.

So it's important that the outputs of g(x) line up with the domain of f(x). Outputs of g(x) must be valid inputs of f(x).

7 0
3 years ago
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