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Mathematics 7 Online
OpenStudy (anonymous):

can you help me match these/@zimmah http://scde.mrooms.org/file.php/1738/Images/5.5HW_M1P.PNG

OpenStudy (anonymous):

OpenStudy (anonymous):

it's asking for login and password

OpenStudy (anonymous):

do u see it now i uploaded it

OpenStudy (anonymous):

heres the answers too it

OpenStudy (anonymous):

\[\large \sqrt[3]{64} = \sqrt[3]{2*32}=\sqrt[3]{2*2*16}=\sqrt[3]{2^6}=4\]

OpenStudy (anonymous):

the answers do not match the questions

OpenStudy (anonymous):

hold rite quick hold up

OpenStudy (anonymous):

\[\sqrt{(-2)^2}=2\]

OpenStudy (anonymous):

heres the questions

OpenStudy (anonymous):

heres the answers

OpenStudy (anonymous):

to solve roots, you have to break the number under the root up in prime factors

OpenStudy (anonymous):

\[\large \sqrt[5]{-243}\] so we have to split -243 up in prime factors. we begin by trying to divide by 2, which won't work because it's an odd number, so we try 3, which works, and makes 81 81 can again be divided by 3 to make 27 27 can again be divided by 3 to make 9 and 9 makes 3 so -243 = -3*3*3*3*3

OpenStudy (anonymous):

can u match the questions with the answers like u did before because im still confused

OpenStudy (anonymous):

\[\sqrt[5]{-3*3*3*3*3}\] there's a 3 in there 5 times, and since it's the 5th degree root we can therefore remove 4 of the 3's and put the 5th 3 in front. so we have \[3\sqrt[5]{-1}\]

OpenStudy (anonymous):

still confused

OpenStudy (anonymous):

so number 3 is question 6

OpenStudy (anonymous):

\[\sqrt{2*2}=2 \] \[\large \sqrt[3]{2*2*2}=2\]

OpenStudy (anonymous):

k i have that one how about the other ones

OpenStudy (anonymous):

the bigger the root, the more of the same number you need to 'come out' of the root

OpenStudy (anonymous):

i am not trying to feed you answers i'm trying to make you understand how it works

OpenStudy (anonymous):

i am

OpenStudy (anonymous):

first you divide the big number under the root into the smallest bits you can

OpenStudy (anonymous):

than you count how many of the same number you have, and see if you have enough of any number to break free of the root

OpenStudy (anonymous):

if you have the 5th root, you need 5 of the same number, if you have the 3rd root, you only need 3. etc.

OpenStudy (anonymous):

so i would do=4096/4?

OpenStudy (anonymous):

you divide by prime numbers always

OpenStudy (anonymous):

k so im doing it rite

OpenStudy (anonymous):

it's best to just go through prime numbers in order, starting with the smallest. so first 2, then 3, then 5, then 7, then 11, then 13, then 17 and you'll probably won't need any higher

OpenStudy (anonymous):

well you can divide by 4 but that's basically dividing by 2 twice

OpenStudy (anonymous):

so is the answer four cause i divied and got 1024

OpenStudy (anonymous):

and then i counted

OpenStudy (anonymous):

\[\large \sqrt[4]{-4096} = \sqrt[4]{4*1024}\]

OpenStudy (anonymous):

oh and the - sign

OpenStudy (anonymous):

but you can still keep dividing, it's still not in the smallest form

OpenStudy (anonymous):

keep dividing until you can't divide any of the numbers under the root anymore

OpenStudy (anonymous):

you can probably divide it a lot of times by 2

OpenStudy (anonymous):

is it +13?

OpenStudy (anonymous):

look how many times can you divide 4096 by 2?

OpenStudy (anonymous):

2 times

OpenStudy (anonymous):

really?

OpenStudy (anonymous):

4

OpenStudy (anonymous):

4096 / 2 = 2048 2048/2 = 1024 that's 2 times, but i guess we can do that about 10 more times or so

OpenStudy (anonymous):

so its 2

OpenStudy (anonymous):

.....

OpenStudy (anonymous):

i'm done for today

OpenStudy (anonymous):

=[

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