Algebra
posted by Marysvoice .
Multiplying sq rts
sqrt18a^7b times sqrt27a^8b^6
Jake 1214
18 = 9 * 2 then sqrt 18 = 3sqrt2
sqrt a^7 = a^3 * sqrt a
27 = 9 * 3 then sqrt 27 = 3 sqrt3
sqrt a^8 = a^4
sqrt b^6 = b^3
Now just multiply the liketerms together.
3sqrt6 and sqrt2a^4?

.. 3sqrt6 and sqrt2a^4?
So you simplified the problem first as I said above and you get:
3(a^3)sqrt(2a) * 3(a^4)(b^3)sqrt(3)
and then you multiply those 2 things together and you get:
9(a^7)(b^3)sqrt(6a)
Respond to this Question
Similar Questions

math calculus please help!
l = lim as x approaches 0 of x/(the square root of (1+x)  the square root of (1x) decide whether: l=1 or l=0 or l=1 Let me make sure I understand the question. Do we have lim x>0 x/[sqrt(1+x)  sqrt(1x)] ? 
Inequality
When I solve the inquality 2x^2  6 < 0, I get x < + or  sqrt(3) So how do I write the solution? 
Math(Roots)
sqrt(24) *I don't really get this stuff.Can somebody please help me? 
math,algebra,help
Directions are simplify by combining like terms. x radiacal 18 3 radical 8x^2 can someone show me how to do these types of problems. thanks I cant determine the second term. For the first, I think you meant x sqrt(18) which reduces … 
Math
How do you find a square root of a number that's not a perfect square? 
Mathematics
sqrt 6 * sqrt 8 also sqrt 7 * sqrt 5 6.92820323 and 5.916079783 So you can see the steps — sqrt 6 * sqrt 8 = sqrt 48 sqrt 7 * sqrt 5 = sqrt 35 I hope this helps a little more. Thanks for asking. 
Math Help please!!
Could someone show me how to solve these problems step by step.... I am confused on how to fully break this down to simpliest terms sqrt 3 * sqrt 15= sqrt 6 * sqrt 8 = sqrt 20 * sqrt 5 = since both terms are sqrt , you can combine … 
Calculus
Please look at my work below: Solve the initialvalue problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/ Sqrt(4^24(1)(6)))/2(1) r=(16 +/ Sqrt(8)) r=8 +/ Sqrt(2)*i, alpha=8, Beta=Sqrt(2) y(0)=2, e^(8*0)*(c1*cos(0)+c2*sin(0))=c2=2 … 
Math/Calculus
Solve the initialvalue problem. Am I using the wrong value for beta here, 2sqrt(2) or am I making a mistake somewhere else? 
Algebra
Evaluate sqrt7x (sqrt x7 sqrt7) Show your work. sqrt(7)*sqrt(x)sqrt(7)*7*sqrt(7) sqrt(7*x)7*sqrt(7*7) sqrt(7x)7*sqrt(7^2) x*sqrt 7x49*x ^^^ would this be my final answer?