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Bogdan [553]
3 years ago
6

SOMEONE PLEASE JUST ANSWER THIS FOR BRAINLIEST!!!

Mathematics
1 answer:
Zanzabum3 years ago
8 0

20c(4 in the top right)d(7 in the top right)

this is a guess so take that with what you will

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Which quadratic equation has no real roots?
n200080 [17]

Answer:

A and D

Step-by-step explanation:

By finding the roots of all the equations, we see that options A and D are the only ones that require an imaginary number in order to solve

3 0
4 years ago
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I<br> rep<br> What is the solution of 4(2y + 1) = 2(y - 13)<br> What is the answer
Novosadov [1.4K]

Answer:

- 5

Step-by-step explanation:

Step 1:

4 ( 2y + 1 ) = 2 ( y - 13 )

Step 2:

8y + 4 = 2y - 26

Step 3:

6y + 4 = - 26

Step 4:

6y = - 30

Answer:

y = - 5

Hope This Helps :)

3 0
3 years ago
The coordinates (-5,2) represent which point?
PSYCHO15rus [73]

Answer:

d

Step-by-step explanation:

3 0
2 years ago
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AaBbCc x AaBbCc (use for all 3 questions) 1. What is the probability that this individual will: AABBcc 2. What is the probabilit
tankabanditka [31]

Answer: 1. P = 1/64

             2. P = 1/32

             3. P = 1/8

Step-by-step explanation:

In genetics, an ofspring inherits one copy of an allele of each parent, which can be described in the tables below called Punnett Square:

For crossing Aa x Aa:

       A      a

A    AA   Aa

a    Aa    aa

For crossing Bb x Bb:

        B      b

B     BB    Bb

b     Bb    bb

For crossing Cc x Cc:

       C      c

C    CC    Cc

c     Cc     cc

We can separate them because they are assorted independently.

For offspring with <u>genotype</u> <u>AABBcc</u>, probability will be:

P(AA) = 1/4

P(BB) = 1/4

P(cc) = 1/4

As all three probabilities has to happen at the same time, it is a "E" rule:

P(AABBcc) = (\frac{1}{4}) (\frac{1}{4}) (\frac{1}{4})

P(AABBcc) = 1/64

Probability for the individual to be AABBcc is 1/64 or 1.56%.

<u>Genotype</u> <u>AaBBcc</u>:

P(Aa) = 2/4 = 1/2

P(BB) = 1/4

P(cc) = 1/4

P(AaBBcc) = (\frac{1}{2}) (\frac{1}{4}) (\frac{1}{4})

P(AaBBcc) = 1/32

Probability for the individual to be AaBBcc is 1/32 or 3.12%

<u>Genotype</u> <u>AaBbCc</u>:

P(Aa) = 1/2

P(Bb) = 1/2

P(Cc) = 1/2

P(AaBbCc) = (\frac{1}{2}) (\frac{1}{2}) (\frac{1}{2})

P(AaBbCc) = 1/8

Probability for the individual to be AaBbCc is 1/8 or 12.5%.

4 0
3 years ago
Which of the following numbers have 16 as a factor<br> 96, 144, 218, 276, 304?
Komok [63]
The answer is the last one 
6 0
3 years ago
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