All categories

3 years ago

Find: (4x2y3 + 2xy2 – 2y) – (–7x2y3 + 6xy2 – 2y)

Mathematics

1 answer:

Ivahew [28]3 years ago

8 0

Answer:

10x2y3 -4xy2

Step-by-step explanation:

Just take the like terms and separate them-

1) 4x2y3 - (-7x2y3)= 10x2y3

2)2xy2 - 6xy2= -4xy

3)-2y - (-2y)= 0y

Hope this helps :)

You might be interested in

Reduce sixteen over twenty four

yawa3891 [41]

16/24 = 4/6 which = 2/3

5 0

3 years ago

Read 2 more answers

Ben is taking a typing class, and the following line graph represents the number of words per minute he was able to type each we

Crank

During week 2, he could type 15 words per minute, at his fourth week, he could type 30. So 30-15=15. He could type 15 words faster on week 4 than he could at week 2.

3 0

4 years ago

Tony plans to deposit $1,000 at the end of each of the next three years. if his funds earn 5% compounded annually, how much w

coldgirl [10]

Answer= $632.98 in compunded interest

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

Can someone help me with this

Kisachek [45]

Answer:

2, 4, 6, and 8, respectively

Step-by-step explanation:

The rule says that 2 times the x value, which we are given, equals to y, which we must find. So all we hve to do is multiply each of the given x values by two to get the y value.

7 0

3 years ago

Read 2 more answers