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liberstina [14]
3 years ago
6

Determine the domain of the function graphed above.

Mathematics
2 answers:
hichkok12 [17]3 years ago
8 0
B. Urhrherhrhhejjrheheheueiejebe
kobusy [5.1K]3 years ago
5 0

Answer:

the domain of given f is (-2,4)

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Laboratory tests show that the lives of
defon

Answer:

34%

Step-by-step explanation:

LammettHash is right just take it as a whole number (for those of you using acellus)

6 0
3 years ago
What is the next number in the pattern?<br> 1, 4, 7, 10,_
Firlakuza [10]

Answer:

13

Step-by-step explanation:

Throughout this pattern, each previous number is adding 3.

Hope this helps! :)

4 0
3 years ago
Find the value of x. Write your answer in simplest form.
givi [52]

Answer:

X = 8

Step-by-step explanation:

Pythagorean triplate

8,8,8√2

4 0
3 years ago
Read 2 more answers
There is an unknown number •it is a four digit number •the tens are double the ones •the thousands are double the tens •the sum
oksano4ka [1.4K]

Answer:

8542

Statement

1.It is a four digit number

2.The tens are double the ones

3.The thousands are double the tens

4.The sum of the digits is nineteen

Step-by-step explanation:

It is a four digit number

lets take 4 digit number-ABCD

The tens are double the ones

C=2D-----------------------------1

The thousands are double the tens

A=2C-------------------------------2

According to equation 1 and 2

A=4D

If we take

D=1,2

Possible number

4B21 OR 8B42

The sum of the digits is nineteen

I use hit and trial method

4B21- If I take B is maximum

4+9+2+1\neq 19

8B42- If, i try to find value of B

8+B+4+2=19---------------------3

14+B=19\\B=19-14\\B=5

So the unknown number is-  8542

6 0
4 years ago
In how many ways can 2 red, 2 black, 3 white and 2 blue balls be selected from 4 red, 3 black, 4 white and 8 blue balls? In how
Montano1993 [528]
From the total pool of colored balls, one can choose 2 reds, 2 blacks, 3 whites, and 2 blues in

\dbinom42\cdot\dbinom32\cdot\dbinom43\cdot\dbinom82=6\cdot3\cdot4\cdot28=2016

ways.

I'm assuming no ball of the same color is distinguishable from any other ball of the same color. So when I'm considering the possible arrangements, if I had lined up the ball as

red1 - black - red2 - ...

then this would be no different that

red2 - black - red1 - ...

So I now have 9 balls to arrange, which means there are 9!=362,880 total possible permutations of them. But order among distinct colors is assumed to not matter. This means I have to divide the total number of permutations by the number of ways I could permute balls of the same color. Then there would be a total of

\dfrac{9!}{2!\cdot2!\cdot3!\cdot2!}=7,560

ways of arranging the balls I had selected.
5 0
3 years ago
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