How do i calculate: ((6 x y−2 x y)−x2y2)+10 x y

2 answers:

5 0

6xy−2xy−x2y2+10xy=6xy+−2xy+−x2y2+10xy=6xy+−2xy+−x2y2+10xy=(−x2y2)+(6xy+−2xy+10xy)=−x2y2+14xy
=−x2y2+14xy

miss Akunina [59]3 years ago

4 0

4xy-x2y2=-10xy
-x2y2=-14xy
x2y2=14xynd class

i think its the answer it can be solved after but it will only get complicated ....it depends on yor chapter a

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a gardener has 8 pounds of fertilizer. He packs the fertilizer into 5/8 pound bags. What is the maximum number of 5/8 pound bags

navik [9.2K]

A gardener has 8 pounds of fertilizer. He packs the fertilizer into 5/8 pound bags. What is the maximum number of 5/8 pound bags he can pack?

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3 years ago

Simplify x^3y^4/3y^4

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3 years ago

Read 2 more answers

There are 10 true-false questions and 20 multiple choice questions from which to choose a five-question quiz how many ways can t

beks73 [17]

Answer:

In 68229 ways can the quiz be selected such that there is atleast three multiple choice questions

Step-by-step explanation:

Given:

Number of True or false questions= 10

Number of multiple choice questions= 20

To Find:

How many ways can 5 questions can be selected if there must be at least three multiple choice questions =?

Solution:

Combination

A combination is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of the selection does not matter. In combinations, you can select the items in any order.

The question States there sholud be ATLEAST 3 multiple choice question,

So, we may have

(3 Multiple choice question and 2 true or false question) or

(4 Multiple choice question and 1 true or false question) or

(5 Multiple choice question and 0 true or false question)

Required Number of ways = (20C3 X10C2) +(20C4 X10C1) + (20C5 X10C0)

Required Number of ways $=(\frac{20!}{20!(20-3)!}\times\frac{10!}{10!(10-2)!})+(\frac{20!}{20!(20-4)!} \times \frac{10!}{10!(10-1)!}) +(\frac{20!}{20!(20-4)!} \times \frac{10!}{10!(10-0)!})$