Question

can someone HHHHHHHHHHHHHEEEEEEEEELLPP
MATH911!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

tofu on the case!

Answer 2

Where are u stuck>

Answer 3

\[6\sqrt{5000}-\sqrt{288}+\sqrt{32}\]

Answer 4

does anyone have an answer PLEASE

Answer 5

Are you looking for an explanation or an answer?

Answer 6

sure

Answer 7

yes i am

Answer 8

Notice that
5000 = 5 x 1000
= 5 . 10^3
= 5 . 2^3 . 5^3
= 2^3 . 5^4

hence

sqrt(5000) = sqrt( 2^3 . 5^4)
= sqrt (2^2) . sqrt(5^4) . sqrt(2)
= 2 . 5^2 . sqrt(2)
= 50 sqrt(2)

Now find similar expressions for sqrt(288) and sqrt(32). I.e., write each of them as something times sqrt(2).

Then you can just add and subtract these terms.

So what are sqrt(288) and sqrt(32) in simplified terms?

Answer 9

\[5000=2\times 2500\] so
\[\sqrt{5000}=\sqrt{2500\times 2}=\sqrt{2500}\times \sqrt{2}=50\sqrt{2}\]

Answer 10

Looks like JamesJ got it covered..

Answer 11

thanks anyway

Answer 12

similarly
\[\sqrt{228}=\sqrt{144\times 2}=12\sqrt{2}\] and
\[\sqrt{32}=\sqrt{16\times 2}=4\sqrt{2}\]

Answer 13

@sat73: it's my sense that students at this level don't have a good feel for these higher perfect squares, so I prefer using prime factorizations; but of course what you're writing is correct.

Answer 14

put it all together and you get
\[300\sqrt{2}-12\sqrt{2}+4\sqrt{2}\] etc
\[

Answer 15

i want to ask another question

Answer 16

pretty sophisticated math for fist grade

Answer 17

*first

Answer 18

@jamesj you may be right, but i have noticed that it is usually easier for people to see a perfect square like 144 and 16 than it is to factor a number into the product of primes. at least for smallish numbers

Answer 19

Find the area and perimeter of the rectangle. Write each answer in simplest form.