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Kamila [148]
3 years ago
14

The dimensions of a patio are shown above what is the area in square feet of the patio?​

Mathematics
1 answer:
Verizon [17]3 years ago
5 0

Answer:

uhh the picture is sideways .

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4(c + d) -3c + 5d when c=1.5 and d= -6.2
patriot [66]

Answer:

-54.3

Step-by-step explanation:

4(1.5−6.2)−4.5−31

=(4)(−4.7)−4.5−31

=−18.8−4.5−31

=−23.3−31

=−54.3

7 0
2 years ago
The table shows the relationship between a, the area of a rectangle, and h, its height, when the base remains constant.
olganol [36]

Answer:

a = 4h.

Step-by-step explanation:

Area (a) = base (b) * height (h).

When h = 2, a = 8:

a =  bh

6 = 2b

b = 4

The result 4 is the same for  h = 5, 7 and 12.

So the required equation is

a = 4h.

4 0
3 years ago
SUPER EASY WILL GIVE BRAINLEIST IF DONE BEFORE 10 PM PLS HURRY. IF U HAVE TIME PLS HELP ME ON OTHER QUESTIONS THANK YOU <3
Murljashka [212]
Tell me if you have any questions or you need a step by step explanation

8 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
Rafael bought a bag of candy that contains 50 pieces. 30 of those pieces4
ludmilkaskok [199]

tell rafel to be careful , he might get diabetes

Step-by-step explanation:

8 0
3 years ago
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