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ladessa [460]
3 years ago
9

What is the GCF of 84 and 56

Mathematics
2 answers:
Elden [556K]3 years ago
8 0
Prime factorization for both numbers:

84 = 2*2*3*7

56= 2*2*2*7

prime factors: 2*2*7

answer: 28
vfiekz [6]3 years ago
6 0

Answer:

The GCF is 28

Step-by-step explanation:

<u>Step 1:  Find the factors of both numbers</u>

84 → 1,  2 ,  3 ,  4 ,  6 ,  7 ,  12 ,  14 ,  21 ,  28 ,  42 ,  84

56 → 1 ,  2 ,  4 ,  7 ,  8 ,  14 ,  28 ,  56

The factor that is the biggest common in both is the GCF.  We see that 28 is the biggest factor that matches for both numbers.

Answer:  The GCF is 28

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Using the Addition Method, solve for x in the following system of linear equations.
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if 2y + x = 4

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3 years ago
If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar
Oksana_A [137]

Answer:

n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}

Step-by-step explanation:

Lets divide it in cases, then sum everything

Case (1): All 5 numbers are different

 In this case, the problem is reduced to count the number of subsets of cardinality 5 from a set of cardinality n. The order doesnt matter because once we have two different sets, we can order them descendently, and we obtain two different 5-tuples in decreasing order.

The total cardinality of this case therefore is the Combinatorial number of n with 5, in other words, the total amount of possibilities to pick 5 elements from a set of n.

{n \choose 5 } = \frac{n!}{5!(n-5)!}

Case (2): 4 numbers are different

We start this case similarly to the previous one, we count how many subsets of 4 elements we can form from a set of n elements. The answer is the combinatorial number of n with 4 {n \choose 4} .

We still have to localize the other element, that forcibly, is one of the four chosen. Therefore, the total amount of possibilities for this case is multiplied by those 4 options.

The total cardinality of this case is 4 * {n \choose 4} .

Case (3): 3 numbers are different

As we did before, we pick 3 elements from a set of n. The amount of possibilities is {n \choose 3} .

Then, we need to define the other 2 numbers. They can be the same number, in which case we have 3 possibilities, or they can be 2 different ones, in which case we have {3 \choose 2 } = 3  possibilities. Therefore, we have a total of 6 possibilities to define the other 2 numbers. That multiplies by 6 the total of cases for this part, giving a total of 6 * {n \choose 3}

Case (4): 2 numbers are different

We pick 2 numbers from a set of n, with a total of {n \choose 2}  possibilities. We have 4 options to define the other 3 numbers, they can all three of them be equal to the biggest number, there can be 2 equal to the biggest number and 1 to the smallest one, there can be 1 equal to the biggest number and 2 to the smallest one, and they can all three of them be equal to the smallest number.

The total amount of possibilities for this case is

4 * {n \choose 2}

Case (5): All numbers are the same

This is easy, he have as many possibilities as numbers the set has. In other words, n

Conclussion

By summing over all 5 cases, the total amount of possibilities to form 5-tuples of integers from 1 through n is

n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}

I hope that works for you!

4 0
3 years ago
3)June 2019-In the diagram below of AABC, D is a
Alexeev081 [22]

Answer:

BC = 23.8

Step-by-step explanation:

See the diagram attached.

Given AC ║ DE and BD = 5, DA = 12 and BE = 7.

We have to find BC.

Since, AC ║ DE, so, Δ ABC and Δ DBE are similar.

If two triangles are similar then the ratio of their corresponding sides remains the same.

Hence, \frac{BD}{BA} = \frac{BE}{BC}

⇒ \frac{5}{12 + 5} = \frac{7}{BC}

⇒ BC = \frac{7 \times 17}{5} = 23.8 (Answer)

7 0
3 years ago
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