All categories

2 years ago

An integer is 5 more than 4 times another. If the product of the two integers is 51, then find the integers.

Mathematics

1 answer:

Yakvenalex [24]2 years ago

7 0

Answer:

3 and 17 are required integers

Step-by-step explanation:

Let one integer be x
And other integer be y

x = 4y + 5

xy = 51

According to question :

➞ (4y + 5)y = 51

➞ 4y² + 5y = 51

➞ 4y² + 5y - 51 = 0

➞ 4y² + 17y - 12y - 51 = 0

➞ y (4y + 17) - 3(4y + 17) = 0

➞ (y - 3)(4y - 17) = 0

➞ y - 3 = 0

➞ y = 3

Hence,

➞ y = 3

➞ x = 4y + 5

➞ x = 4 (3) + 5

➞ x = 12 + 5

➞ x = 17

∴ The required integers are 3 and 17

You might be interested in

Help please! This is due today in a few hours!!

avanturin [10]

Perimeter is 40 and area is 60 something. Sorry if I’m wrong I tried to help you tho. But If I am wrong look at other people’s answers

5 0

2 years ago

A square is a quadrilateral always sometimes never

lbvjy [14]

A quadrilateral is a shape with 4 sides, regardless of their lengths, angles, or symmetry.

Therefore, a square is always a quadrilateral.

6 0

3 years ago

Read 2 more answers

Find the surface area of the prism HURRY HURRY

Assoli18 [71]

Answer:

should be 192

Step-by-step explanation:

4 0

2 years ago

Human visual inspection of solder joints on printed circuit boards can be very subjective. Part of the problem stems from the nu

olga_2 [115]

Answer:

The probability that the selected joint was judged to be defective by neither of the two inspectors is $P(A' n B' ) = 0.8855$

The probability that the selected joint was judged to be defective by inspector B but not by inspector A is $P(A' n B) =0.0403$

Step-by-step explanation:

From the question we are told that

The sample size is $n_s = 10000$

The number of outcome for inspector A is $n__{A}} = 742$

The number of outcome for inspector B is $n__{B}} = 745$

The number of joints judged defective by both inspector is $n(A u B) = 1145$

The the probability that the selected joint was judged to be defective by neither of the two inspectors is mathematically represented as

$P(A' n B' ) = 1 - P(A u B)$

Now

$P(A\ u \ B) = \frac{n(Au B)}{n_s }$

substituting values

$P(A\ u \ B) = \frac{1145}{ 10 000 }$

$P(A' n B' ) = 1 - \frac{1145}{10 000}$

$P(A' n B' ) = 0.8855$

the probability that the selected joint was judged to be defective by inspector B but not by inspector A is mathematically represented as

$P(A' n B) = P(A \ u \ B) -P(A)$

Now

$P(A) = \frac{n__{A}}{n_s}$

substituting values

$P(A) = \frac{742}{10 000}$

$P(A' n B) = \frac{1145}{10 000} - \frac{742}{10 000}$

$P(A' n B) =0.0403$

7 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