Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

emma97 Group TitleBest ResponseYou've already chosen the best response.1
\[f(3) \sqrt[3]{x2}\]
 11 months ago

bibby Group TitleBest ResponseYou've already chosen the best response.3
\[f(x)=\sqrt[3]{x2}\] ?
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
\[f(3)=\sqrt[3]{x2}\]
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
i forgot to put an = sign the first time
 11 months ago

bibby Group TitleBest ResponseYou've already chosen the best response.3
are we finding f'(3) either way to get the inverse, let's replace f(x) with y \[y = \sqrt[3]{x2}\] switch x and y \[x = \sqrt[3]{y2}\] cube both sides \[x^3 = {y2}\] \[x^3+2=y\]
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
\[x^{3}+2=y^{3}\] is that right?.. i have no idea I'm so stressed
 11 months ago

bibby Group TitleBest ResponseYou've already chosen the best response.3
How'd you get y^3? Remember that square roots and squares are inverse operations, right? \[(\sqrt{x})^2 = x\] \[(\sqrt[3]{x})^3 = x\]
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
oh i got confused i thought you said for ME to cube both sides lol. I'm sorry I'm extremely tired and have to get a bunch of work done by like tomorrow. ughhh. my bad
 11 months ago

bibby Group TitleBest ResponseYou've already chosen the best response.3
lol no worries, just always go over your work a bazillion times so that you can prove to yourself that your answer is right. ANYHOW, the inverse is now f(x) =x^2+2 f(3) = 3^2+2 = ??
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
wait so the answer is f(x) =x^2+2 ?
 11 months ago

bibby Group TitleBest ResponseYou've already chosen the best response.3
whoops./ you generally say inverse with a little dash f'(x) so the equation for the inverse is f'(x) = x^2 + 2 then you plug in 3 into the inverse equation (assuming that's the question)
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
heres the actual question with the answer choices so you can see
 11 months ago

bibby Group TitleBest ResponseYou've already chosen the best response.3
god damn. I kept saying x^2 +2 but I meant x^3 + 2
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
hahah so is it the 3rd option then?
 11 months ago

bibby Group TitleBest ResponseYou've already chosen the best response.3
it is, it is...
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
thank youuu! i have another question! do you mind helping me??
 11 months ago

bibby Group TitleBest ResponseYou've already chosen the best response.3
are you gonna post it here or make a new question? I don't mind
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
ill post it here! and thanks! (:
 11 months ago

bibby Group TitleBest ResponseYou've already chosen the best response.3
uh you'd multiply by the conjugate. pardon my handwriting
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
i dont understand...
 11 months ago

bibby Group TitleBest ResponseYou've already chosen the best response.3
to simplify radicals you multiply by the conjugate pair which means flipping the sign
 11 months ago

bibby Group TitleBest ResponseYou've already chosen the best response.3
why do you have that answer selected?
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
what do you mean that answer selected?
 11 months ago

bibby Group TitleBest ResponseYou've already chosen the best response.3
what are the answer choices?
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
the ones with the bubble next to them on the link i posted!
 11 months ago

bibby Group TitleBest ResponseYou've already chosen the best response.3
rofl I didn't see it scrolled down a bit.
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
hahahah its all good! (:
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
so what is the answer?
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Perform the operation bibby suggested and then simplify.
 11 months ago

bibby Group TitleBest ResponseYou've already chosen the best response.3
you can use wolfram if you don't like learning http://www.wolframalpha.com/input/?i=%28sqrt%283%29sqrt%286%29%29%2F%28sqrt%283%29%2Bsqrt%286%29%29
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
its not that i don't like learning its that i am in a hurry and i have no idea whats going on.
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{\sqrt{3}\sqrt{6}}{\sqrt{3}+\sqrt{6}} \cdot \frac{\sqrt{3}\sqrt{6}}{\sqrt{3}\sqrt{6}}\] Do this multiplication here.
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Why are you in a hurry? Is this a test?
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
idk how to..
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
i had to go out of town so I'm behind on a bunch of work due like really soon
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
So you don't know how to multiply something like this (a+b)(c+d)?
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
A lot of Americans like to use this thing called the foil method.
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
i have heard of it but idk how to really use it
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
(a+b)(c+d)=ac+ad+bc+bd
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
all you did was distribute those right?
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
dw:1388451041982:dw
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
yepdw:1388451059969:dw
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
(a+b)(c+d)=a(c+d)+b(c+d) =ac+ad+bc+bd
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
So if you had \[(\sqrt{3}\sqrt{6})(\sqrt{3}\sqrt{6})\] what do you have after multiplying?
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
idk how to group square roots together or multiply them together
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Do that foil method then we will work on simplifying
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
hold on its gonna take awhile to plug in
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
You don't have to use what I gave you as a formula. Just distribute. Take each of the terms in the first grouping and multiply to each term in the second grouping
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
oh wow my computer plugged in the wrong numbers so thats not even the right question, this is the right question...
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Well I will show you how to do the first one and you try this one.
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
i tried solving it and i think its the second answer choice
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{\sqrt{3}\sqrt{6}}{\sqrt{3}+\sqrt{6}} \cdot \frac{\sqrt{3}\sqrt{6}}{\sqrt{3}\sqrt{6}}\] This was the operation you were asked to perform. I was having you look at the numerator first. So looking at the numerator if we multiply the top out we would get: \[\frac{\sqrt{3} \sqrt{3}\sqrt{3} \sqrt{6}\sqrt{6} \sqrt{3}+\sqrt{6} \sqrt{6}}{(\sqrt{3}+\sqrt{6})(\sqrt{3}\sqrt{6})}\]
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Do you see how I got that top part?
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
If a, b>=0 then \[\sqrt{a} \cdot \sqrt{b} =\sqrt{a \cdot b} \]
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
so basically \[\sqrt{3} \cdot \sqrt{3} =3 \text{ and } \sqrt{3} \cdot \sqrt{6}=\sqrt{18}=\sqrt{9 \cdot 2} =\sqrt{9} \cdot \sqrt{2}=3\sqrt{2}\]
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
combine like terms on top and multiply the bottom out i have to go for now
 11 months ago

Chooch146 Group TitleBest ResponseYou've already chosen the best response.1
This has to be the longest problem eve
 11 months ago

Chooch146 Group TitleBest ResponseYou've already chosen the best response.1
ever* lol my name is eve:P
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
hahaha hey eve! and yah i know…i have a lot of stuff i need help withhh ughh
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
hahah whats up!
 11 months ago

Chooch146 Group TitleBest ResponseYou've already chosen the best response.1
not mch jst listening to music i was about to leave lol
 11 months ago

emma97 Group TitleBest ResponseYou've already chosen the best response.1
hahaha ok then! lol
 11 months ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.