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kvasek [131]
3 years ago
12

Help me and don't put wrong answer. 30 points.

Mathematics
2 answers:
AveGali [126]3 years ago
8 0

Answer:

House of Scorpion I know this one, It's a Metaphor. And if your looking for the rest of the answers go to quizziz House of scorpion final exam if you want all answers

Step-by-step explanation:

vovangra [49]3 years ago
4 0

Answer:

I believe this is a METAPHOR cause you cant really throw your voice as rope.

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Is 5.75 a terminating decimal or repeating decimal?
yawa3891 [41]
Terminated. Any decimal that repeats goes on forever. Such as pie: 3.1415...
3 0
3 years ago
Solve the following quadratic by factoring.
Lina20 [59]

Answer:

4, -1

Step-by-step explanation:

We can reverse the FOIL (First, outside, inside, last) method to break down the equation. The question is asking to solve by factoring so you want to find the right combination of numbers this equation can be broken down into:

x^2 - 3x - 4 = 0

Because we want x squared, x needs to be multiplied by itself, so we can put x in the first slot for each.

(x + or - ?) ( x + or - ?) = 0

Then we need to find numbers that could be added to get -3 and multiplied to get -4. The only set of numbers that works for this is -4 and 1. Note that the sign you put in front of each number has an impact on your answers. With this we get:

(x - 4) (x + 1) = 0

To test that this is equal to the original equation, simply multiply it out using FOIL.

x * x = x^2

x * 1 = x

x * -4 = -4x

-4 * 1 = -4

Putting each component into an equation:

x^2 + x - 4x - 4 = 0

Simplifying:

x^2 - 3x - 4 = 0

Once we are sure it is still the same equation, we find the solutions. We know 0 multiplied by anything equals 0, so to get 0 as the answer, one of the sets in the parentheses must equal 0. (It doesn't matter what the other one is as long is one equals 0)

Therefore, we have 2 solutions, 4, and -1 because if x is 4, 4-4 is 0 which solves the equation, and if x is -1, then -1 + 1 is 0 which also solves the equation.

You can also check your answers by plugging them back into the original equation.

5 0
3 years ago
Aaron has 16 blue pens and 12 red pens.
Maslowich

Answer:

4

Step-by-step explanation:

  • 16÷4=4 12÷3=4 meaning is.the lowest common multiple
7 0
3 years ago
Subtract -2x^2+4x-1 from 6x^2+3x-9 your answer should be a polynomial in standard form
Sphinxa [80]

the answer -8^2+1x-10

5 0
3 years ago
Read 2 more answers
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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