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MissTica
2 years ago
13

Ntercepts

Mathematics
1 answer:
never [62]2 years ago
5 0
It will be X (-3,0), y(0,-6/5)



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Selectie correct answer.
denis-greek [22]

Answer:

OD.0

Step-by-step explanation:

because 2(X+4)-1 = 2x+7

or, 2x+8-1 = 2x+7

or, 2x+7= 2x+7

and it cancels out eachother so it will be 0

4 0
3 years ago
Solve for x<br> 9x= 1/2(20x-8)
LenKa [72]

Answer:

4

Step-by-step explanation:

9x=10x-4

x=4

3 0
3 years ago
Read 2 more answers
A square room has a tiled floor with 81 square tiles. How many tiles are along an edge of the room?
Strike441 [17]
As there are 81 tiles total, that makes your area 81. Being a square, that makes your area equation A=s^2. As you need a side, this makes your equation s=√A. Plugging in 81 as A, we get s=√81=9, so an edge has 9 tiles.
4 0
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
One third of the students at Finley high school play sports. Two fifths of the students who play sports are girls. What fraction
Cerrena [4.2K]
You would have to do 1/3x2/5 which is 2/15. So 2/15 of the girls in the school play sports. 
7 0
3 years ago
Read 2 more answers
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