At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

u guyz ought to help everyone.......plz help me also.......

Sidd, please stop stalking me.

2) -15 Sqrt[5]
not exactly what the first is asking though

siddharth tiwari : ....... well i need help with these so ? ......

i have no idea polopak , i get so confused with these sqrts :/

Don't worry about the square roots for now. Can you factor 80?

well the common gcf of all 5, 80, 45 is 5

for problem 2

Now try doing the same thing for 45.

And be sure to group up any squares you find

1,5,9,45 = factors for 45

Ignore 1 and itself.

and for 5 , it would be just 5

45 = 9*5, and 9 is what?

3*3

\(3^2\)

Now any square factor we can pull out of the square root by removing the square.

\(7*\sqrt{2^2 * 2^2 * 5} = 7 * \sqrt{2^2} * \sqrt{2^2} * \sqrt{5} = ?\)

i multiply the sqrts ?

No, take the square root of the squared term. What is \(\sqrt{2^2}\)

If a >= 0, then \(\sqrt{a^2} = a\)

Are you still with me?

yes

Ok, great. So what's \(\sqrt{2^2}\)

sqrt. 4

Err
\(\sqrt{3^2} = 3\)

so it would just be 2 and 3 ?

7 * 2 sqrt. 5 ?

correction : 7 * 2 * 2 * sqrt. 5 ?

Right. So now simplify that

7*2*2 = ?

becuase their are 2 [sqrt.2^2]

can we still simplfy that ?

2 sqrt. 5 ?

4 sqrt . 5 *3^2

So what is the next step now that you have that?

Just like the last one, we pull out each squared factor.

4*3 sqrt.5 ?

Yes. Which is?

What's 4*3?

12 sqrt. 5

Right. So now we have \(1\sqrt{5} - 28\sqrt{5} + 12\sqrt{5}\)

-15 sqrt. 5 ?

Yes.

and for the 1st problem , i factor out which is a ?

Go ahead and do it yourself, step by step and I'll point out when/if you get off track.

Just post each step as you go.

What's the first step?

[4^3 sqrt. a] + [^3 sqrt. 64a] : what do i do with the powers before the sqrt ?

are you still here ?

yes i still need help please

polpak: i think it'll result to 16\[^{3} \sqrt{a}\]

What was the first step you did?