Find the product or quotient.
\[(\sqrt[4]{6})(\sqrt[3]{6})\]
Then for my first step I got
Hmm give me some time to solve it out on paper.
\[6\frac{ 1 }{ 4 } * 6\frac{ 1 }{ 3 }\]
hmm have you seen something like \( \large x^2 * x^3 = x^5\) before?
6^(7/12) is your answer I think.
@fall15 I believe EndersWorld has a deeper misunderstanding seeing that they thought 6 1/4 * 6 1/3 would be the second step perhaps if we gave them an explanation, it would help them with their other homework questions as well
@mLe yes I have, but I have a different formula to use, would you like me to give it to you?
|dw:1523918049764:dw|
Sorry took me some time to draw it.
^that looks like how my teacher did it
Do you understand what I did?
Yes! I just need someone to walk through it with me, I’m a little slow
ohhhh right right right I misread that you wrote 6 1/4 as in 6.25 rather than 6^(1/4) as an exponent @ EndersWorld and awesome @ fall15 , I really like how you did that :)
I have a couple I need help with a couple I need y’all to check, do y’all have a few?
|dw:1523918271155:dw|
That should help you to understand how it works. It's a bit messy though. . .
I get it.
Great!
cool, what else can we help you with? :D
I have five questions I need someone to check, and then I need help with 4 others
Can y’all help?
I'll try. Post them.
\[(36)^1/2\]
That’s 36 with a exponent of 1/2
Then I got square root of 36, then 6 as a final answer
Yeah it's correct. Good job ender.
2^1/2 * 32^1/2 square root of 2 * square root of 32 Then I got \[\sqrt{6^4}\] Then got 6 as a final answer once again
hmm it's not right
It's 8
Hmmm...
I’m confused..
Square root of 6 is.... uh... nvm
\(\large \sqrt{2} * \sqrt{32} = \sqrt{2*32}\)
OOOOOO so, square root of 64=8?
You got it!
8^2/3
\[\sqrt[3]{8^2}\]
\[\sqrt[3]{64} =4\]
perfect :)
x^1/6= \[\sqrt[6]{x}\]
Correct.
x^2/7=\[\sqrt[7]{x^2}\]
correct
Are you good with exponential form?
Let me take look. Post them.
Square root of -10 I got 3.9e-7 somehow...
You should be using imaginary numbers. . .
interesting... I don't believe it is possible to get a square root of a negative number (you would get imaginary numbers)
i should be in your answer. . .
Hmm.. oo I got 3.9i the first time... thought I was wrong
The letter i
are you asked to estimate it? I feel like you could probably keep the square root 10
No no no that's wrong. It's intended that way for a purpose.
i= -1
give me a second to draw it all out and post it for you
Straight off my homework paper^
|dw:1523920196164:dw|
YOu could keep the i inside with the 10 but it's not a proper answer that makes sense then so you write the i outside of the square root.
How would that be in exponential form though?
I have one more exponential form question... and it has a variable in it...
|dw:1523920390315:dw|
That's how you would write in exponential form. ^
hmm
but the i is not in the square root
That's right. It shouldn't be.
|dw:1523920507957:dw|
Yeah that's how it's done.
but no the i should be 10 i with the 1/2
:thonk: my head hurts...
|dw:1523920595347:dw|
i = (-1)^(1/2) by itself, so i in the final answer should not be in the parenthesis
\[\sqrt{(7x)^3}\]
One does not simply put THAT in exponential form...^
I'm honestly losing track. . .
it's the same idea \(\large \sqrt{x} = x^{1/2} \) so \(\large \sqrt{(7x)^3} = ( ~(7x)^3~)^{1/2}\)
Me too,... thank you guys so much! <3 sadly I have to go... I guess I’m leaving 3/10 unanswered XD, I got chores to do!
wait, I wasn't done ;-;
tbh most of this stuff is old for me. I barely remember some of it. so angle could be right or wrong. I have no idea. All I know is that there should an i and a 10 with 1/2 as the exponent.
there is \(\large (x^2)^3 = x^{2*3} = x^6\)
so \(\large \sqrt{(7x)^3} = ( ~(7x)^3~)^{1/2}\) becomes \(\large (7x)^{3*(1/2)} = ~?\)
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