factor out perfect squares.
#1. √250h^4k^5 = √25*10*h^4*k^4*k = √(25h^4k^4)*10k = 5h^2k^2 √10k
#2. wrong again. You don't factor them all out
#3. ok
#4. ok
#5. nope. √8*6x = √48x = 4√3x ...
1- Simplify the radical expression: √250h^4k^5
a) hk√125
b) 5√10h^4k^5
c) 5h^2k^2√10k
d) 25hk√10k **
2- Simplify the radical expression: √21y * 5√49y
a) 5y√1,029
b) 5√1,029y^2
c) 35y√21
d) 35√21y^2 **
3- Simplify the radical expression: √20x^13y^5 / 5xy^7 (this whole fraction is a square root, not just the nominator, this is the same for all the fractions in this question with the √ sign)
a) √4x^12 / y^2
b) 2x^6 / y **
c) 2√x^12 / y^2
d) 2x^6y
4- Simplify the radical expression: 5√7 + 2√175
a) 7√182
b) 15√14
c) 7√175
d) 15√7 **
5- Simplify the radical expression by rationalizing the denominator
√8 / √6x
a) 2√3x / 3x
b) 2√2x / x **
c) √6x / 3x
d) 4√3x / 3x
PLS HELP
4 answers
1- 5h^2k^2 √10k
2- 35y√21
3- 2x^6 / y
4- 15√7
5- √6x / 3x
I changed 1, 2, and 5. Are these right? For #2 my original answer was 5√1,029y^2 but I'm pretty sure I got it wrong.
2- 35y√21
3- 2x^6 / y
4- 15√7
5- √6x / 3x
I changed 1, 2, and 5. Are these right? For #2 my original answer was 5√1,029y^2 but I'm pretty sure I got it wrong.
#1,2 ok
For #5, √8 / √6x = √48x / 6x = 4√3x / 6x = 2√3x / 3x
For #5, √8 / √6x = √48x / 6x = 4√3x / 6x = 2√3x / 3x
Oh I see now.
Thanks Mr. oobleck!
Thanks Mr. oobleck!