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svetlana [45]
3 years ago
14

Determine all the real roots of the equation...

Mathematics
1 answer:
Nitella [24]3 years ago
5 0
Let's solve it!

So, let's study each factor.

x + 7 = 0 -> x = -7

x² - 49 = 0 -> x² = 49 - > x = ± √49 -> x = <span>± 7

Final answer: x = 7 and x = -7
</span>
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Alex, Bob, and Claudia split 126 cm wire evenly among themselves. They then proceeded to cut their pieces of wire into smaller,
Westkost [7]
<h3>Answer:</h3>
  • Bob
  • 18 cm
<h3>Step-by-step explanation:</h3>

If Alex cut his wire 18 times, he ended up with 19 equal pieces. He kept 7, so has 7/19 of his 1/3 of the wire.

Bob cut his wire 20 times, so ended up with 21 pieces, of which he kept 9. So he has 9/21 = 3/7 of his 1/3 of the wire.

Claudia kept 1/13 of her 1/3 of the wire, so has the smallest piece.

Bob kept (3/7)·(1/3)·126 cm = 18 cm.

Alex kept (7/19)·(1/3)·126 cm ≈ 15.47 cm.

Bob kept the longest part of the original wire.

3 0
3 years ago
Genetic Inheritance: Calculate probability of two children having the same genotype for achrondoplasia Achondroplasia is a commo
dimaraw [331]

Answer:

The correct answer is - 1/2 or 50% for first and second child to be affected.

Step-by-step explanation:

Achondroplasia is an autosomal dominant disorder. Autosomal dominant disorder refers to the presence of a single copy of the defective gene that is enough to lead to dwarfness.

A cross of achondroplasia (Aa) parent to a person of normal height (aa) result in half of their children will be affected with dwarfism and the other half will be normal.

a  cross between affected or dwarf  and normal parent

     Aa X aa

Punnett square:

         a a

A  Aa Aa

a aa aa

Aa- dwarfness

aa- normal height

The probability that both their first child and second child would have achondroplasia is

2/4 =1/2 or 50%.

6 0
3 years ago
PLLZZZZZ HELPPP!!!!!!
maks197457 [2]
Ms. Harris commission rate is 11.76 percent
7 0
3 years ago
M. Score: 0 of 1 pt
beks73 [17]

Answer:

  3×5×53

Step-by-step explanation:

You can use divisibility rules to find the small prime factors.

The number ends in 5, so is divisible by 5.

  795/5 = 159

The sum of digits is 1+5+9 = 15; 1+5 = 6, a number divisible by 3, so 3 is a factor.

  159/3 = 53 . . . . . a prime number,* so we're done.

795 = 3×5×53

_____

* If this were not prime, it would be divisible by a prime less than its square root. √53 ≈ 7.3. We know it is not divisible by 2, 3, or 5. We also know the closest multiples of 7 are 49 and 56, so it is not divisible by 7. Hence 53 is prime.

5 0
3 years ago
A bag contains 3 red, 4 blue, and 6 yellow marbles. One marble is selected at a time, and once a marble is selected, it is not r
astraxan [27]

The probability of choosing two yellow marbles = 5/26

Step-by-step explanation:

Step 1 :

Given

Total number of marbles in the bag = 13 marbles

Number of yellow marbles = 6.

We choose 2 marbles from the bag without replacing and need to determine the probability that both are yellow.

Step 2 :

Probability of finding yellow marble in the first try = Number of yellow marbles / Total number of marbles = \frac{6}{13}

Once the yellow marble is selected we have 5 yellow marbles among 12 remaining marbles in the bag.

Probability of finding yellow marble in the second try = Number of yellow marbles / Total number of marbles = \frac{5}{12}

Step 3:

P(yellow in first 2 tries) = P(yellow in first try) * P(yellow in second try)

P(yellow in first 2 tries) = \frac{6}{13} * \frac{5}{12} = \frac{5}{26}

Step 4:

Answer:

The probability of choosing two yellow marbles = 5/26

4 0
3 years ago
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