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same idea: distribute then combine sqrt's
im not sure how i would but it in distubutive property
5 srqt 3 (2 sqrt3 sqrt 3 + 6 sqrt 3 sqrt 19)??
in letters: A * (B+C) = A*B + A*C take the "stuff" in front (in this case 3 sqrt3 ) and multiply it times each thing inside
5 sqrt 3 (2 sqrt 3 + 6 sqrt 19) you have two terms inside the parens: 2 sqrt3 and 6 sqrt19 you multiply each term by 5 sqrt3 for the first term just write it like this 5*sqrt3 * 2 * sqrt3 ( I put in * to show the multiplication, but in algebra you normally don't show the multiplication sign, and just know it is there.) however, put the numbers first (when you multiply you can switch the order) 5*2 * sqrt3 * sqrt3 you know what 5*2 is , right ? do you know what sqrt3 * sqrt3 is ? and do the same thing for the 2nd term: 6 sqrt19
ok so 10 sqrt 6 and 5 * sqrt 3 * 6 * sqrt 19 giving me 5 * 6 and sqrt 3 * sqrt 19 then giving me 30 sqrt 57 Answer = 10 sqrt 6 + 30 sqrt 57
giving me 5 * 6 and sqrt 3 * sqrt 19 then giving me 30 sqrt 57 that part is correct. do the first part again 5*2 * sqrt3 * sqrt3
5*2 = 10 sqrt 3 * sqrt 3 = sqrt 9 10 sqrt 9
yes. but what is the square root of 9 ?
yes so 10*sqrt(9) is 10*3 or 30
so right now im looking at D for my answer
what is the first term? (5*2 * sqrt3 * sqrt3) + (5 * sqrt 3 * 6 * sqrt 19)
the first term is 10 sqrt 9 simplified its 10 sqrt 3
next term is 30 sqrt 57
the first term is 10 sqrt 9 simplified its 10 sqrt 3 <-- no. sqrt(9) is 3 square root of 9 is 3 because 3*3=9
so if you see sqrt(3*3) you know that is 3
so, the answer for this question is 10 sqrt 9 simplified its 10 *3 =30 for the first term.
can you get the answer ?
yes, also I noticed a typo in my rules: sqrt(x) * sqrt(x) = x is the rule or sqrt( x*x) = x or sqrt(x^2) = x <-- (it is x, not 2 like I had) those are 3 different ways of saying the same thing. try to remember this rule.
ok thanks i'll post the last question if ur ready