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beks73 [17]
2 years ago
12

Factor -10x + 2 please answer!

Mathematics
1 answer:
MaRussiya [10]2 years ago
8 0

Answer:

Step-by-step explanation:

you just give 2 a common factor

2(-5x+1)

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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
Two cars leave the same parking lot, with one heading north and the other heading east.
Otrada [13]

Answer:

wouldn't it be 3 kilometers because 3 this way plus 3 that way = 6 away from each other?

8 0
3 years ago
Take different values for x and find the value of 3x+5.​
Eva8 [605]

Answer:

when x=1

3x+5.=3×1+5=8

when x=0

3x+5.=3×0+5=5

Step-by-step explanation:

when x=-1

3x+5.=3×-1+5=-3+5=2

4 0
2 years ago
Read 2 more answers
A hula hoop has an circumference 150.72 cm. What is the area of the hula hoop? (round to the nearest hundredth)
Leya [2.2K]

Answer:

Area = 1808.64 cm

Step-by-step explanation:

Circumference of a circle = 2πr

Circumference = 150.72

π = 3.14

r = radius

150.72 = 2 x 3.14 x r

150.72 = 6.28 r

r = 150.72/6.28

r = 24cm

Radius = 24

Area of a circle = πr²

Area = 3.14 x 24²

Area = 3.14 x 576

Area =1808.64 cm

6 0
2 years ago
What are the leading coefficient, constant term and degree, if any, of the algebraic expression
nasty-shy [4]
Coefficients are the numbers in from with the variable, the constant is just a number and the degree is the exponent.
6 0
3 years ago
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