haltiamali
  • haltiamali
(2a^3b^2) (b^3)^2 Simplify. (The ^'s mean to the power of..)
Mathematics
katieb
  • katieb
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ajprincess
  • ajprincess
\(\large (x^y)^z)=x^{y*z}\) using this can u simplify \(\large (b^3)^2\)?
haltiamali
  • haltiamali
So it would be \[b^6..?\]
ajprincess
  • ajprincess
ya very good:)

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ajprincess
  • ajprincess
\(\large x^y*x^z=x^{y+z}\) Usin this can u simplify \(\large b^2*b^6\)?
haltiamali
  • haltiamali
Is the answer \[2a^5b^12\]
haltiamali
  • haltiamali
that last number is supposed to be a 12
ajprincess
  • ajprincess
nope. \(\large x^y*x^z=x^{y+z}\) using this hint can u simplify \(\large b^2*b^6\)?
haltiamali
  • haltiamali
\[b^8?\]
ajprincess
  • ajprincess
ya very good:)
haltiamali
  • haltiamali
\[2a^3b^8\] is the answer?
ajprincess
  • ajprincess
yup:)
haltiamali
  • haltiamali
thx. :) can you help me with the other 3?
ajprincess
  • ajprincess
ya sure:)
haltiamali
  • haltiamali
\[(-4xy)^3 (-2x^2)^3\]
ajprincess
  • ajprincess
\(\Large (-2x^2)^2=(-2)^2*(x^2)^2\) What is \(\Large (-2)^2\)?
haltiamali
  • haltiamali
4
ajprincess
  • ajprincess
ya good:) What do u get when u simplify \(\Large (x^2)^2\)?
haltiamali
  • haltiamali
4x?
ajprincess
  • ajprincess
nope \(\Large (a^b)^c=a^{b*c}\) using this simplify \(\Large (x^2)^2\)
haltiamali
  • haltiamali
ohh. \[x^4\] right?
ajprincess
  • ajprincess
ya:) \(\Large(-4xy)^3=(-4)^3*(x^1)^3*(y^1)^3\) Nw can u simplify (-4)^3*(x^1)^3*(y^1)^3\)
haltiamali
  • haltiamali
-64 x^4 y^4?
haltiamali
  • haltiamali
oh wait
haltiamali
  • haltiamali
hahaha I get it :D
haltiamali
  • haltiamali
\[(-14xy^4) (-1/2xy^3)\]
haltiamali
  • haltiamali
that's negative half
haltiamali
  • haltiamali
Would it be \[-14 1/2 x+y^7?\]
ajprincess
  • ajprincess
wait a sec. oops sorry it is \(\Large (-2x^2)^3\) right? So \(\Large (-2x^2)^3=(-2)^3*(x^2)^3\) We r nt yet done with our previous problem. so it will be better if we finish that and come to this one. What is \((-2)^3)? What do u get when u simplify \((x^2)^3\)?
haltiamali
  • haltiamali
\[(-2x^6) \] right?
ajprincess
  • ajprincess
\(\Large (x^2)^3=x^6\) that's right:) but what is \(\Large (-2)^3=?
haltiamali
  • haltiamali
oh -8
ajprincess
  • ajprincess
ya:) so \(\Large (-2x^2)^3=-8x^6\) Nw \(\Large(-4xy)^3=(-4)^3*(x^1)^3*(y^1)^3\) What is \(\Large (-4)^3=\)? What do u get when u simplify \(\Large (x^1)^3\) and \(\Large (y^1)^3\)?
haltiamali
  • haltiamali
-64 x^4 and y^4
ajprincess
  • ajprincess
\(\Large(a^b)^c=a^{b*c}\) usin this general rule only u have to simplify \((x^1)^3\) and \((y^1)^3\)
haltiamali
  • haltiamali
oh. x^3 and y^3
ajprincess
  • ajprincess
ya good:)
ajprincess
  • ajprincess
\(\Large(-4xy)^3=-64x^3y^3\) \(\Large(-2x2)^3=-8x^6\) \(\Large(-64x^3y^3)*(-8x^6)=(-64*8)*(x^3*x^6)*y^3\) To simplify \(\Large(x^3*x^6)\) u have to use the general rule \(\Large(a^b)*(a^c)=a^{b+c}\)
ajprincess
  • ajprincess
Can u figure out the final answer @haltiamali
haltiamali
  • haltiamali
yes :) tysm
ajprincess
  • ajprincess
k nw moving on to ur next problem \[(-14xy^4)(\frac{ -1 }{ 2 }xy^3)=(-14*-\frac{ 1 }{ 2 })(x^1*x^1)*(y^4*y^3)\] See if u can simplify nw. If u find any difficulties while solving pls do ask:)
haltiamali
  • haltiamali
Thanks ! :)
ajprincess
  • ajprincess
Welcome:)

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