All categories

3 years ago

How do you simplify this.

Formula1" title=" \sqrt{24x ^{2} } y ^{3} " alt=" \sqrt{24x ^{2} } y ^{3} " align="absmiddle" class="latex-formula">

Mathematics

1 answer:

Ad libitum [116K]3 years ago

8 0

$\bf \sqrt{24x^2y^3}~~ \begin{cases} 24=2\cdot 2\cdot 2\cdot 3\\ \qquad 2^2\cdot 6\\ y^3=y^{2+1}\\ \qquad y^2\cdot y^1 \end{cases} \\\\\\ \sqrt{2^2\cdot 6\cdot x^2\cdot y^2\cdot y^1}\implies 2xy\sqrt{6y}$

You might be interested in

A salesperson's commission rate is 5 %. What is the commission from the sale of 32,000 worth of furnaces? Use pencil and pape

Len [333]

Answer: Answer:

Its sad that you have to uses this app to get the right answers.

Have you ever stopped to ponder if someone is trying to scam/Catfish you?

Also I bet that you haven't even tried reading this question.

I know that you can do it yourself!

If not then, Sorry to hear.

Also

The answer is:

Do it yourself, Because if you try..You'll have a better answer then I ever could. Or anyone TBH

6 0

3 years ago

Chris can type 490 words in 7 min how many works can he type PER min <br> You get 14 points

amm1812

<h2>Answer: 70 words per min</h2><h2 /><h2>Step-by-step explanation: We just divided 490 by 7 to get how many words he can type for 1 min!</h2>

5 0

3 years ago

How can you decompose the composite figure to determine its area?

laiz [17]

see the attached figure to better understand the problem

we know that

The Area of the composite figure is equal to the sum of Area 1, Area 2 and Area 3

The Area 1 is a triangle

The Area 2 is a rectangle

The Area 3 is equal a semicircle

therefore

<u>the answer is the option</u>

a triangle, a rectangle, and a semicircle

6 0

3 years ago

Read 2 more answers

Please answer this question correctly and i'll mark you as brainlilest! Thank you.

raketka [301]

Answer:

$\frac{ {7}^{4} ( - 2) {}^{4} {11}^{3} 6}{7 {}^{3} {2}^{2} {6}^{2} 11 {}^{2} } = \frac{7 \times {2}^{2} \times 11}{6} = \frac{308}{6} = \frac{154}{3} \\ \\ \frac{ {7}^{0} {12}^{0} {17}^{0} }{ {57}^{0} + ( {42}^{0} \times ( {13}^{0} ))} = \frac{1}{1 + 1} = \frac{ 1}{2}$

4 0

3 years ago

Let AA and BB be events such that P(A∩B)=1/73P(A∩B)=1/73, P(A~)=68/73P(A~)=68/73, P(B)=21/73P(B)=21/73. What is the probability

krok68 [10]

Answer:

$P(A \cup B) = \frac{5}{73} +\frac{21}{73} -\frac{1}{73}=\frac{25}{73}$

Step-by-step explanation:

Let A and B events. We have defined the probabilities for some events:

$P(A') =\frac{68}{73} , P(B) =\frac{21}{73} , P(A \cap B) =\frac{1}{73}$

Where A' represent the complement for the event A

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: $P(A)+P(A') =1$

So for this case we can solve for P(A) like this:

$P(A) = 1-P(A') = 1-\frac{68}{73}=\frac{5}{73}$

And now we can find $P(A \cup B)$ using the total probability rul given by:

$P(A \cup B) = P(A)+P(B) -P(A \cap B)$

And if we replace the values given we got:

$P(A \cup B) = \frac{5}{73} +\frac{21}{73} -\frac{1}{73}=\frac{25}{73}$

And that would be the final answer.

5 0

3 years ago