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2 years ago

Find the HCF of x^4 y^2 and x^3 y^3

Mathematics

1 answer:

forsale [732]2 years ago

8 0

Here's your answer down here↓:

Step-by-step explanation:

x^3 - y^3 = (x - y)(x^2 + xy + y^2)

(x^4 - y^4) = (x^2 - y^2)(x^2 + y^2) = (x - y)(x + y)(x^2 + y^2)

Ok. So the factor (x-y) appears once in the top line and once in the second line. So we are going to take it the least amount of times.

So, the factor (x + y) appears in the top line zero times and in the second line one time so we will take it where it appears the least which is zero times so we are still at (x - y)

And it Same goes for the factors of (x^2 + y^2) and (x^2 + xy + y^2)

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HELP ASAP! please need answer fast

Ray Of Light [21]

14^15/14^5= 14^10

This is because the 14^5 in the denominator cancels out. If it helps the equation can be written out as so:
(14^10 * 14^5)/ 14^5

Final answer: D

6 0

3 years ago

Read 2 more answers

(x-1)^2+(2-x)^2-3(x-5)*(x-5)

Natali5045456 [20]

Answer:

$-x^2+24x-70$

Step-by-step explanation:

$(x-1)^2+(2-x)^2-3(x-5)*(x-5)\\=> (x - 1)^2 + (2 - x)^2 - 3(x - 5)^2\\=> x^2 - 2x + 1 + 4 - 4x + x^2 - 3(x - 5)^2\\=> x^2 - 2x + 5 - 4x + x^2 - 3(x - 5)^2\\=> x^2 - 6x + 5 + x^2 - 3(x - 5)^2\\=> 2x^2 - 6x+5 - 3(x - 5)^2\\=> 2x^2 - 6x+5 - 3(x^2 - 10x + 25)\\=> 2x^2 - 6x+5 - 3x^2+30x-75\\=> -x^2 - 6x + 5 + 30x - 75\\=> -x^2 + 24x + 5-75\\=> -x^2+24x-70$

Formula used =

$(a-b)^2 = a^2+ 2ab + b^2$

8 0

3 years ago

Help me plz thank u

Svetlanka [38]

Answer:

Step-by-step explanation:

2x+6=5x-9

3x=15,

x=5

2(5)+6=10+6=16

5(5)-9=25-9=16

16+16=32

8 0

3 years ago

Read 2 more answers

PLEASE NEED HELP WITH HOMEWORK Write the correct postulate, theorem, property, or definition that justifies the statement below

tatuchka [14]

Answer:

Congruent angles have equal measure.

4 0

2 years ago

The probability that a randomly selected 3-year-old male chipmunk will live to be 4 years old is 0.96516.

mezya [45]

Using the binomial distribution, it is found that there is a:

a) The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

b) The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

c) The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

-----------

For each chipmunk, there are only two possible outcomes. Either they will live to be 4 years old, or they will not. The probability of a chipmunk living is independent of any other chipmunk, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

0.96516 probability of a chipmunk living through the year, thus $p = 0.96516$

Item a:

Two is P(X = 2) when n = 2, thus:

$P(X = 2) = C_{2,2}(0.96516)^2(1-0.96516)^{0} = 0.9315$

The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

Item b:

Six is P(X = 6) when n = 6, then:

$P(X = 6) = C_{6,6}(0.96516)^6(1-0.96516)^{0} = 0.80834$

The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

Item c:

At least one not living is:

$P(X < 6) = 1 - P(X = 6) = 1 - 0.80834 = 0.19166$

The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

A similar problem is given at brainly.com/question/24756209

6 0

3 years ago