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olga_2 [115]
3 years ago
11

1.) The _____ value of your ordered pair is the input

Mathematics
2 answers:
Sergeu [11.5K]3 years ago
8 0
1.) X-value

Reason: the set of inputs, also called x-coordinates, of the order pairs.

2.) Y-value

Reason: the range the set of outputs, also called y-coordinates of ordered pairs.
torisob [31]3 years ago
3 0
1. x-value
2. y-value
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Y=2-5x

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y= .40 - x
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Find the additive inverse or opposite of -4/3
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A little difficult question for me, it is multiple choice answer but my teacher taught this before the break. If anyone could he
Sever21 [200]
<h3>Answer: Choice D</h3>

Divide both sides of the first equation by 7, then add the result to the second equation

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

Explanation:

We can multiply both sides of an equation by the same number and the equation will be equivalent to the original.

For example, if we had  x = 5, then we could get 2x = 10 after multiplying both sides by 2. The reason they are equivalent equations is because the same x value is the solution for both equations.

We can also divide both sides of an equation by the same number and the two equations would be equivalent. We can go from 2x = 10 back to x = 5 when we divide both sides by 2.

----------

If we divide both sides of 7x - 21y = 14 by 7, then we end up with x - 3y = 2. Simply divide each term (7x,  -21y, and 14) by 7.

Because 7x-21y=14 and x-3y=2 are equivalent, this means we can replace the "7x-21y=14" with "x-3y=2"

--------

The new system would be

x-3y = 2

2x+3y = 11

From here you add straight down. Doing so will have the y terms add to -3y+3y = 0y = 0. After this point the y terms are eliminated and you can solve for x just like with any other equation of one variable.

7 0
3 years ago
Of the 9-letter passwords formed by rearranging the letters AAAABBCCC (4 A's, 2 B's, and 3 C's), I select one at random. Determi
Tanya [424]

Answer:

a) 3

b) (8!/9!)-(7!/9!)

c) (1-(8!/9!))*(7!/9!)

Step-by-step explanation:

a)With 4 As ;  2Bs and 3Cs it is possible to get a palindrome if you fixed the  letters C according to: (2) in the extremes of the word and the other one at the center therefore you only have palindrome in the following cases

<u>C</u> (       ) <u>C</u> (       ) <u>C</u>

To fill in the gaps we have  4 letters  A and 2 letters B, wich we have two divide in two palindrome gaps,  

AAB         and    BAA the palindrome is  C  AAB C BAA C

BAA         and    AAB    "           "           is  C  BAA C AAB C  

ABA         and    ABA    "           "           is  C  ABA C ABA C

b) 4 A  ;   2B  ; 3C

We have the total number of elements  9, so the total number of possible outcomes is : 9!

Total events: 9!

if we fixed 3 C we have (the group of 3 Cs becoming one element) so the total amount of events with 3 adjacent Cs is: 7!

Therefore the probability of having 3 adjacent Cs is: 7!/9!

If we fixed only 2 Cs we have:

4 A  ; 2 B  ; 2C  : 1C

Total number of words (events) in this case is 8! (2C becomes 1 element)

so the total numbers of events is 8! the probability in this case is 8!/9!(this value includes cases of adjacent 3 Cs previous calculated ) so this value minus the case of 3 adjacent Cs ) give us 2 adjacent C and the other no next to them

Probability (of words with 2 adjacent Cs and the other no next to them is); 8!/9! - 7!/9!

c) Probability of B apart from each other is the whole set of events minus those where 2 B are adjacent or (become 1 element)

4 A ; 2B ; 3C

Total of events 9! and events with adjacent B is 8!/9!

Therefore the probability of words with 3 adjacent Cs and 2 B separeted is

the probability of 3 adjacent Cs (7!/9!) times probability of words with no adjacent Bs wich is (1-(8!/9!))*(7!/9!)

5 0
3 years ago
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