4√18+√32-√8
The answer is?
Add or subtract. simplify by collecting like radical terms, if possible.?
2009-03-24 3:25 pm
回答 (7)
2009-03-24 3:35 pm
✔ 最佳答案
4√18+√32-√8=4√(9*2)+√(16*2)-√(4*2)
=12√2+4√2-2√2
=14√2 answer//
參考: check my answer here:
http://www.myalgebra.com/algebra_solver.aspx
2016-12-01 2:12 pm
nicely ... because of the fact the quantity the quantity in the novel sign is the comparable for all of the words , you in ordinary terms could upload the numbers exterior the novel..... so the respond on your question would be ..... 13 sq. root of three.
2009-03-24 3:42 pm
4√18+√32-√8 = (4*3√2 + 4√2) - 2√2
= (12√2 + 4√2) - 2√2
= 16√2 - 2√2
= 14√2
= (12√2 + 4√2) - 2√2
= 16√2 - 2√2
= 14√2
2009-03-24 3:41 pm
4√18 + √32 - √8
= 4√(3^2 * 2) + √(4^2 * 2) - √(2^2 * 2)
= 4√(3^2) * √2 + √(4^2) * √2 - √(2^2) * √2
= 4(3)√2 + 4√2 - 2√2
= 12√2 + 4√2 - 2√2
= (12 + 4 - 2)√2
= 14√2
= 4√(3^2 * 2) + √(4^2 * 2) - √(2^2 * 2)
= 4√(3^2) * √2 + √(4^2) * √2 - √(2^2) * √2
= 4(3)√2 + 4√2 - 2√2
= 12√2 + 4√2 - 2√2
= (12 + 4 - 2)√2
= 14√2
2009-03-24 3:38 pm
12√2 + 4√2 - 2√2 =14√2
2009-03-24 3:34 pm
4 sqrt(18) = 4 sqrt(9*2) = 3*4 sqrt(2) = 12 sqrt(2)
sqrt(32) = sqrt(16*2) = 4 sqrt(2)
sqrt(8) = sqrt(4*2) = 2 sqrt(2)
12 sqrt(2) + 4 sqrt(2) - 2 sqrt(2) = 14 sqrt(2)
sqrt(32) = sqrt(16*2) = 4 sqrt(2)
sqrt(8) = sqrt(4*2) = 2 sqrt(2)
12 sqrt(2) + 4 sqrt(2) - 2 sqrt(2) = 14 sqrt(2)
2009-03-24 3:33 pm
20 sqrt 2
收錄日期: 2021-05-01 12:11:23
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