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2 years ago

Can somebody help me plz with number 13,14,15,16,17

Mathematics

1 answer:

yuradex [85]2 years ago

5 0

13. There are 3 feet in a yard. You divide 34 by 3 and get 11 with one left over. So it should be 11 yards and 1 foot.
14. 16 oz. = 1 lb. If you multiply 16 (oz) x 8 (lbs.) you get 128 ounces.

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Find the sum of the series. ∞ 5(−1)nπ2n + 1 32n + 1(2n + 1)! n = 0

wlad13 [49]

Could you give the question more clear?

5 0

2 years ago

Hilda was simplifying some numerical expressions and made each of the following sequences of calculations. Name the mathematical

Galina-37 [17]

Answer:

Problem 1:

5 · (−4/3) · (2/5)

= (−4/3) · 5 · (2/5) Conmutative Property of Multiplication

= (−4/3) · (5 · (2/5)) Associative Property of Multiplication

= (−4/3) · (2/1) = −8/3 = −2*2/3 Multiplied fractions and extracted common factor

Problem 2:

17 + 29 + 3+ 1

= 17 + 3 + 29 + 1 Conmutative Property of Addition

= (17 + 3) + (29 + 1) Associative Property of Addition

= 20 + 30 Added groups

= 50 Added terms

Step-by-step explanation:

For the Problem 1:

In the first step, Hilda applied the Conmutative Property of Multiplication, because she changed the order of the numbers in the product

In the second step, she applied the Associative Property of Multiplication, because she agrouped the product of 5 and 2/5 to perform it sepparately

In the third step, she calculated the product of the fractions -4/3 and 2/1, then she extracted 2 as a common factor to express the fraction as -2*2/3

For the Problem 2:

In the first step, Hilda applied the Conmutative Property of Addition, because she changed the order of the numbers in the sum

In the second step, she applied the Associative Property of Addition, because she associated the addition of 17 and 3 and the addition of 29 and 1, to calculate them in groups.

4 0

2 years ago

Read 2 more answers

[Supporting Standard: I can find the sum of an infinite series.]

Tanya [424]

Answer:

787.5 is the answer.

Step-by-step explanation:

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please Mark as brainlist

4 0

2 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.