is anyone good with rational exponents??
yes
maybe
ahh can you help me with this equation: 3^1/3*9^1/3
probably
factor out the 1/3 because they are the same exponent so you have (3*9)^1/3 which is like the cube root of 27 which is 3
i think the answer is 3sq rt of 3?
i dont know tht
THANK YOU dancr_on_fire!! :)
no problem... i always find it easier to think of factional exponents like roots
(3.3.3)^1/3 = 3, good job dancr
how would i simplify 8^4/3
cube root of 8 is 2, 2^4 = 16
okay awesome!! what about...(8a^-6)^-2/3
with this you multiply your exponents first
multiply the exponents first: \[(8a^{-6})^{-\frac{2}{3}}=8^{-\frac{2}{3}}a^4\]
satellite73 you are the best! are you good with finding LCM?
of polynomials
in english \[8^{-\frac{2}{3}}\] means the reciprocal of the cube root of 8 squared. the cube root of 8 is 2. 2 squared is 4. and the reciprocal of 4 is \[\frac{1}{4}\] so answer is \[\frac{a^4}{4}\]
example?
okay, 400x^2-4y^2, 20x^2+2^yx
do you really mean lcm? because these are sums and difference. ok factor each one. \[400x^2-4y^2=4(100x^2-y^2)\]which is the difference of two squares. \[=4(10x-y)(10x+y)\]
\[20x^2+2y^2=2x(10x+y)\]
so for your least common multiple you need all the factors you see, not repeated!
the factors are \[4\] \[x\] \[10x-y\] \[10x+y\]
okayyy i see! that makes sense!
multiply together to get \[4x(10x+y)(10x-y)\] hope it is clear. same way to find lcm of two numbers.
can you help me with writing this next one in radical form? 6x^3/2
yes that makes way better sense now!
\[x^{\frac{m}{n}}=\sqrt[n]{x^m}\]
for a fractional exponent, numerator is the power, denominator is the root.
\[6x^{\frac{3}{2}}=6\sqrt{x^3}\]\]
check out the paper i sent should make it clear.
okay lemme check it out! thank you!
are you always online? cause you are the biggest help! hopefully you can help me again soon!
i am often here these days practicing my latex. if you want you can email me at satellite73.openstudy@gmail.com
perfect! i am trying to finish an online math class by tomorrow in order to be able to graduate...i have about 70% more to complete! so youre a big help!
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