Mathematics 33 Online OpenStudy (anonymous):

how would you simplify a cubed root radical of 25 times the radical of 125 OpenStudy (anonymous):

ok, you have 25^(1/3) * 125^(1/2)? OpenStudy (anonymous):

yes OpenStudy (anonymous):

ok, so how are 25 and 125 related? OpenStudy (anonymous):

Both divisible by 5 and 25? OpenStudy (anonymous):

ok, cool, so 25 = 5^2 and 125 = 5^2 OpenStudy (anonymous):

5^3, sorry OpenStudy (anonymous):

oh okay yeah OpenStudy (anonymous):

$\sqrt{25} \times \sqrt{125} = 25^{1/3} \times 125^{1/2} = 5^{2/3} \times 5^{3/2} = 5^{4/6} \times 5^{9/6} = 5^{13/6}$ Feel free to correct me, this is my first time using this equation maker. OpenStudy (anonymous):

omg it's too long for the page HAHA OpenStudy (anonymous):

what is the equation maker? OpenStudy (anonymous):

all right OpenStudy (anonymous):

the answer is 5^{13/6} sorry it's too long haha OpenStudy (anonymous):

yeah, that's how to do it easily hrwhyhry OpenStudy (anonymous):

okay.. how on earth did you get that though? OpenStudy (anonymous):

so you have 25=5^2, and 125=5^3....so you have (5^2)^(1/3) * (5^3)^(1/2) OpenStudy (anonymous):

you can then add the exponents...here: $(5^2)^(1/3)$ OpenStudy (anonymous):

ohhhhh oka OpenStudy (anonymous):

y mn OpenStudy (anonymous):

(5^2)^(1/3) --> the square (2) and the (1/3) and now add together, 2/3 OpenStudy (anonymous):

okay thank you so much! OpenStudy (anonymous):

so, you got it? OpenStudy (anonymous):

i think soooo OpenStudy (anonymous):

now how would you do x-$\sqrt{3}\div \sqrt{12}$ OpenStudy (anonymous):

sorry its supposed to be the x- the 3 one OpenStudy (anonymous):

???? OpenStudy (anonymous):

just a second OpenStudy (anonymous):

thank you :) OpenStudy (anonymous):

$(x-\sqrt{3})/\sqrt{12}$ OpenStudy (anonymous):

is that the problem? is it equal to something? OpenStudy (anonymous):

no thats the problem. You only have to simplify it OpenStudy (anonymous):

ok OpenStudy (anonymous):

so what is sqrt(12)? OpenStudy (anonymous):

what are factors of 12? OpenStudy (anonymous):

the square root of 4 times the square root of 3? OpenStudy (anonymous):

right on, so sqrt(4) is simple, sqrt(3) is good because the problem has 3^(1/3) OpenStudy (anonymous):

is the answer 1 over x^1/6 times 2? OpenStudy (anonymous):

1/(x^1/6 * 2)? OpenStudy (anonymous):

yes OpenStudy (anonymous):

$(x-(3^{1/3}))/(2*(3^{1/2}))$ OpenStudy (anonymous):

i had that... i just dont kmow if u can get rid of the three OpenStudy (anonymous):

so you can separate x and -3^(1/3) because they have the same denominator and they are part of a subtraction operation OpenStudy (anonymous):

ok... OpenStudy (anonymous):

thank u for the help by the way OpenStudy (anonymous):

first look at 3^(1/3) / (2 * (3^1/2)) OpenStudy (anonymous):

np OpenStudy (anonymous):

can you simplify that? focus on the 3's OpenStudy (anonymous):

u can divide 3 by 3 and get one right OpenStudy (anonymous):

when you divide by exponents you and subtract the exponents from values with the same base OpenStudy (anonymous):

so 3^(1/3) / 3^(1/2) is 3^(1/3) * 3^-(1/2)...can you can add these exponents --> 1/3 = 2/6, 1/2 = 3/6, need to have similar fractions....so 2/6 - 3/6 = -1/6 OpenStudy (anonymous):

so you have 3^(-1/6) OpenStudy (anonymous):

which is 1/(3^1/6) OpenStudy (anonymous):

yes and u have to bring it below the division sign since it is negative?> OpenStudy (anonymous):

yeah, like in the last message OpenStudy (anonymous):

is the answer x over 3^1/6 *2 OpenStudy (anonymous):

wait, x-1 OpenStudy (anonymous):

(x-1)/(2*(3^1/6)) OpenStudy (anonymous):

$(x-1)/(2*(3^1/6))$ OpenStudy (anonymous):

$(x-1)/(2*(3^{1/6}))$ OpenStudy (anonymous):

how's that look? OpenStudy (anonymous):

Thank you sooo much! I think thats right.....

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