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WARRIOR [948]
2 years ago
13

Simplify. 6a2−4b2+4abc−9b2−4a2+8abc

Mathematics
2 answers:
Arte-miy333 [17]2 years ago
7 0

Answer:

\huge\boxed{\sf 2a^2-13b^2+12abc}

Step-by-step explanation:

<h3><u>Given expression:</u></h3>

=6a^2-4b^2+4abc-9b^2-4a^2+8abc\\\\Combine \ like \ terms\\\\=6a^2-4a^2-4b^2-9b^2+8abc+4abc\\\\=2a^2-13b^2+12abc\\\\\rule[225]{225}{2}

Elena-2011 [213]2 years ago
4 0

Answer:

Hello! The answer is: 2a^2 - 13b^2 + 12abc

Step-by-step explanation:

6a^2 - 4b^2 + 4abc - 9b^2 - 4a^2 + 8abc

Rearrange to combine like terms:

6a^2 - 4a^2 - 4b^2 - 9b^2 + 4abc + 8abc

Combine like terms:

2a^2 - 13b^2 + 12abc

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(-0.4)^2 sum pls i need this by tonight i have to finish my math corrections pls
netineya [11]

Answer:

(-0.4)²

= -0.4×-0.4

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so, 0.4×0.4

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5 0
2 years ago
An experiment has three possible mutually exclusive outcomes, A, B, or C. The odds of A occurring are 3 to 10 and the
Arte-miy333 [17]

Using the relation between odds and probability, it is found that:

1. P(C) = 17/39.

2. The odds of C occurring are 17 to 22.

<h3>What is a probability and what is an odd?</h3>

  • A probability is given by the <u>number of desired outcomes divided by the number of total outcomes</u>.
  • An odd is the number of <u>desired outcomes divided by the number of non-desired outcomes</u>.

Considering the odds, the probabilities for A and B are given as follows:

  • P(A) = 3/(3 + 10) = 3/13.
  • P(B) = 1/(1 + 2) = 1/3.

The sum of all probabilities is of 1, hence the probability of C is found as follows:

\frac{3}{13} + \frac{1}{3} + P(C) = 1

\frac{9 + 13 + 39P(C)}{39} = 1

39P(C) = 17

P(C) = 17/39.

17 desired outcomes and 39 - 17 = 22 non-desired, hence:

The odds of C occurring are 17 to 22.

More can be learned about the relation between odds and probability at brainly.com/question/25683609

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6 0
2 years ago
URGENT !!!!!!!!! Please answer correctly !!!!! Will be marking Brianliest !!!!!!!!!!!!!!!!
babunello [35]

Answer:

slope is 3/5

Step-by-step explanation:

-10-2/-16-4

12/20

6/10

3/5

6 0
3 years ago
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