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kupik [55]
3 years ago
8

How do you work out y+y in algebra

Mathematics
2 answers:
goldenfox [79]3 years ago
8 0
Y+y = 2y there is no other solution.
melisa1 [442]3 years ago
4 0
Y+y=2y as there is two y’s
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"Suppose that allele D produces long toes and allele d produces normal toes. A heterozygous male with the Dd genotype mates with
VARVARA [1.3K]

Answer:

30%

Step-by-step explanation:

The crossing of a male with genotype Dd with a female with genotype dd, yields the following possible outcomes:

Dd, dD, dd, dd.

Therefore, there is a 2 in 4, (50%) chance that their offspring will have the dominant allele. Since its penetrance is 60%, the dominant trait will only manifest itself in 60% of heterozygous individuals. Therefore, the probability that their offspring will have long toes is:

P = 0.50*0.60\\P=30\%

The probability is 30%.

7 0
3 years ago
Using the given rectangular box, find the following.... PLEASE HELP ME ASAP!! SUCK AT MATH!!!
Rasek [7]
I believe it would be 482 mm^2 but I could be wrong

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3 years ago
Solve for x<br><br>4x+6-7+9=18​
Alla [95]

here is your ans have a nice day

6 0
3 years ago
Read 2 more answers
2 Points<br> Which of the following is(are) the solution(s) to . x+6 = 2x - 1?
seropon [69]

Answer:

x = 7

Step by step explanation:

<em>minus</em><em> </em><em>x</em><em> </em><em>on</em><em> </em><em>both</em><em> </em><em>sides</em>

6 = x -1

<em>add</em><em> </em><em>1</em><em> </em><em>on</em><em> </em><em>both</em><em> </em><em>sides</em>

<em>x</em><em> </em><em>=</em><em> </em><em>7</em>

7 0
3 years ago
Let $s$ be a subset of $\{1, 2, 3, \dots, 100\}$, containing $50$ elements. how many such sets have the property that every pair
Tamiku [17]

Let A be the set {1, 2, 3, 4, 5, ...., 99, 100}.

The set of Odd numbers O = {1, 3, 5, 7, ...97, 99}, among these the odd primes are :

P={3, 5, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}

we can count that n(O)=50 and n(P)=24.

 

 

Any prime number has a common factor >1 with only multiples of itself.

For example 41 has a common multiple >1 with 41*2=82, 41*3=123, which is out of the list and so on...

For example consider the prime 13, it has common multiples >1 with 26, 39, 52, 65, 78, 91, and 104... which is out of the list.

Similarly, for the smallest odd prime, 3, we see that we are soon out of the list:

3, 3*2=6, 3*3=9, ......3*33=99, 3*34=102.. 

we cannot include any non-multiple of 3 in a list containing 3. We cannot include for example 5, as the greatest common factor of 3 and 5 is 1.

This means that none of the odd numbers can be contained in the described subsets.

 

 

Now consider the remaining 26 odd numbers:

{1, 9, 15, 21, 25, 27, 33, 35, 39, 45, 49, 51, 55, 57, 63, 65, 69, 75, 77, 81, 85, 87, 91, 93, 95, 99}

which can be written in terms of their prime factors as:

{1, 3*3, 3*5, 3*7, 5*5,3*3*3, 3*11,5*7, 3*13, 2*2*3*3, 7*7, 3*17, 5*11 , 3*19,3*21, 5*13, 3*23,3*5*5, 7*11, 3*3*3*3, 5*17, 3*29, 7*13, 3*31, 5*19, 3*3*11}

 

1 certainly cannot be in the sets, as its common factor with any of the other numbers is 1.

3*3 has 3 as its least factor (except 1), so numbers with common factors greater than 1, must be multiples of 3. We already tried and found out that there cannot be produced enough such numbers within the set { 1, 2, 3, ...}

 

3*5: numbers with common factors >1, with 3*5 must be 

either multiples of 3: 3, 3*2, 3*3, ...3*33 (32 of them)

either multiples of 5: 5, 5*2, ...5*20 (19 of them)

or of both : 15, 15*2, 15*3, 15*4, 15*5, 15*6 (6 of them)

 

we may ask "why not add the multiples of 3 and of 5", we have 32+19=51, which seems to work.

The reason is that some of these 32 and 19 are common, so we do not have 51, and more important, some of these numbers do not have a common factor >1:

for example: 3*33 and 5*20

so the largest number we can get is to count the multiples of the smallest factor, which is 3 in our case.

 

By this reasoning, it is clear that we cannot construct a set of 50 elements from {1, 2, 3, ....}  containing any of the above odd numbers, such that the common factor of any 2 elements of this set is >1.

 

What is left, is the very first (and only) obvious set: {2, 4, 6, 8, ...., 48, 50}

 

<span>Answer: only 1: the set {2, 4, 6, …100}</span>

8 0
3 years ago
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