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anonymous
 one year ago
Which one is greater? =)
anonymous
 one year ago
Which one is greater? =)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1 }{ \sqrt{1} }+\sqrt{\frac{ 1 }{ 3 }}\] or \[\frac{ 1 }{ \sqrt{1} }+\sqrt{\frac{ 1 }{ 4 }}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i got the second one as bigger

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1which one is big, 1/3 or 1/4 ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1would you like to have 1/3 of cheesecake or 1/4 of cheesecake ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but 1/3 is bigger lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait but can u take a look at what i did

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437674327047:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1we can't cancel like that \[\dfrac{a+b}{b}~~\ne~~a\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but doesnt it become a 1?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what does it become then?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1437674587867:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then whats the proper way to do it?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1you have said earlier that 1/3 is greater than 1/4 therefore \(\frac{ 1 }{ \sqrt{1} }+\sqrt{\frac{ 1 }{ 3 }}\) is greater than \(\frac{ 1 }{ \sqrt{1} }+\sqrt{\frac{ 1 }{ 4 }}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but can you solve it out? I need to know why....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0like how do the solutions compare?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1im not getting any ideas, not seeing any neat/standard way for problems like these @freckles

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok can u help me with another problem then?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it's an easy question but i don't know why this website is giving a wrong answer....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If \[(x+3)^{2}=25\] which of the following could be the value of x? a. 8 b. 5 c. 2 d. 5 e. 8

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i thought it was this at first dw:1437675183398:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then i thought it was this : dw:1437675254689:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but the correct answer is 8... and i don't know why

freckles
 one year ago
Best ResponseYou've already chosen the best response.2@ganeshie8 oh how to compare those two numbers?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hi freckles for the first problem I presented can you solve it out for me? because i know that 1/3 is larger but i need to know how the problem is completely solved and compare the two solutions....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I gues i did it the wrong way but fail to understand why it's wrong

freckles
 one year ago
Best ResponseYou've already chosen the best response.2first of all you really did mean: \[ \frac{ 1 }{ \sqrt{1} }+\sqrt{\frac{ 1 }{ 3 }} \text{ or } \frac{ 1 }{ \sqrt{1} }+\sqrt{\frac{ 1 }{ 4 }} \] just asking because I find that they put sqrt(1) weird you know because that is 1 and also find it weird they put 1/sqrt(1) because that is also 1 \[1+\sqrt{\frac{1}{3}} \text{ ? } 1 +\sqrt{\frac{1}{4}} \\ \text{ subtract 1 on both sides } \sqrt{\frac{1}{3}} ? \sqrt{\frac{1}{4}} \\ \text{ squaring both sides } \frac{1}{3} ? \frac{1}{4} \\ \\ \text{ well we know } 4>3 \text{ so } \frac{1}{4} < \frac{1}{3} \\ \text{ which means } \sqrt{\frac{1}{4}} < \sqrt{\frac{1}{3}} \\ \text{ which means } 1+\sqrt{\frac{1}{4}} < 1 +\sqrt{\frac{1}{3}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that still wasn't solved tho =(

freckles
 one year ago
Best ResponseYou've already chosen the best response.2i guess I'm asking what are you looking for?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0like the problem solved out

freckles
 one year ago
Best ResponseYou've already chosen the best response.2to me I showed why sqrt(1/4)<sqrt(1/3) since 4>3 which means 1/4<1/3 taking square root of both sides the direction of inequality still holds

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmmm okay i gotcha can you help me with the second problem please?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i just really fon't understand why it's 8 instead of 2 or 8

freckles
 one year ago
Best ResponseYou've already chosen the best response.2dw:1437675933396:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.2so you are right it is 2 or 8

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait what? is this for all square root problems? it's either a 5 or 5?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh okay but why is it positive 8 then?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[x^2=a \text{ if } a \text{ is positive then yes } x=\sqrt{a} \text{ or } x=\sqrt{a} \\ \text{ since both } (\sqrt{a})^2=a \text{ and } (\sqrt{a})^2=a\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I guess I could also include 0 and negative a but x^2=0 only gives x=0 and x^2=a where a is negative gives complex solutions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so you're saying it can be 2 and 2 and also 8 and 8?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[(x+3)^2=25 \text{ gives us } x+3=5 \text{ or } x+3=5 \] is what I'm saying

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\text{ which gives } x=53 \text{ or } x=53 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so positive 8 is wrong?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2yeah if the problem really is (x+3)^2=25 because if x=8 then (8+3)^2=(11)^2=121 121 is not 25

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ahhhhh okay i gotcha ^_^ thank youuuuuuuuuuuuuu!!~~~

freckles
 one year ago
Best ResponseYou've already chosen the best response.2can we go back to the previous problem @yomamabf were you expecting a certain method on that one?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well i wanted to have it solved out so i know that the end solution would coincide with the rule that the smaller denominator will always produce a larger number

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437676640564:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0like i don't understand why thats wrong

freckles
 one year ago
Best ResponseYou've already chosen the best response.2these look different then the other two values you gave

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{1}{\sqrt{1}}+\sqrt{\frac{1}{4}} \text{ or } \frac{1}{\sqrt{1}}+\sqrt{\frac{1}{3}} ?\] are these not the two values you were comparing ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437676741343:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea those are but i just solved it is why they look different

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{1}{\sqrt{1}}+\frac{\sqrt{1}}{\sqrt{3}} \\ \] So you tried to combine fractions : This means you want to multiply the first fraction by sqrt(3)/sqrt(3) \[\frac{\sqrt{3}}{\sqrt{3}}+\frac{1}{\sqrt{3}} \\ \frac{\sqrt{3}+1}{\sqrt{3}} \text{ but you cannot say } \frac{a+b}{a} =\frac{\cancel{a}+b}{\cancel{a}}=1+b \\ \text{ but you can say} \frac{a+b}{a}=\frac{a}{a}+\frac{b}{a}=1+\frac{b}{a}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you can "cancel out" common factors in division

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh okay so u can cancel out if you divide but cant reduce in this situation because we're not dividing?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2fraction bar is division

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so when can i cancel them out ? because I remember doing that before for another problem and it was the right thing to do.... =( so confused

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\text{ say you have this } \\ \frac{\sqrt{3}+2 \sqrt{3}}{\sqrt{3}} \\ \text{ the \top and bottom share a common factor } \sqrt{3} \\ \ \text{ that is you can factor out a } \sqrt{3} \text{ on \top } \\ \frac{\sqrt{3}(1+2)}{\sqrt{3}} \\ \text{ notice top and bottom have } \sqrt{3} \text{ factor \in common } \\ \text{ we know } \frac{\sqrt{3}}{\sqrt{3}} =1 \text{ so we can do } \\ \frac{\cancel{\sqrt{3}}(2+1)}{\cancel{3}}=\frac{2+1}{1}=2+1=3\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Like for this one dw:1437677209526:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.2basically all terms have the have the same factor for example: \[\frac{6x^3+5x^2+x}{2x^2x}\] all the terms on top have a common factor x all terms on bottom have a common factor x you can divide each term by x \[\frac{6x^2+5x+1}{2x1}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea i understand that but for roots its a bit different right?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{ 6 (\sqrt{3})^3+5 (\sqrt{3})^2+\sqrt{3}}{2 (\sqrt{3})^2\sqrt{3}} \text{ divide top and bottom by } \sqrt{3} \\ \text{ divide each term in the top by } \sqrt{3} \\ \text{ divide each term in the bottom by } \sqrt{3} \\ \frac{6 (\sqrt{3}))^2+5 \sqrt{3}+1}{2 \sqrt{3}1}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hey for the problem u put on top aren't u supposed to solve it like this? dw:1437677469609:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437677555931:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{\sqrt{3}+1}{\sqrt{3}} \\ \text{ if you did \choose to try what you did earlier } \\ \text{ you would also need to divide the } 1 \text{ by } \sqrt{3} \\ \text{ that is you would have } \\ \frac{\frac{\sqrt{3}}{\sqrt{3}}+\frac{1}{\sqrt{3}}}{ \frac{\sqrt{3}}{\sqrt{3}}} =\frac{1+\frac{1}{\sqrt{3}}}{1}=1+\frac{1}{\sqrt{3}} \text{ but this is what you had to \begin with } \\ \text{ so you were working backwards to what you started with }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you could but another way is the way I did

freckles
 one year ago
Best ResponseYou've already chosen the best response.2and the way I did it was to give you an example of when you can cancel

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if it involves addition and subtraction don't reduce it but if it involves multiplication ad division then reduce it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh wait nevermind HAHAHAHAHHAHA OHHHHHHHHHHHHHHHHHHH IM SO STUPID OKAY I GOT IT

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0SORRY hahahaha okay i got it lmaoooooo

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i see what i did wrong this is what it's supposed to be dw:1437677816464:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you so much for breaking everything down you're the best <33333

freckles
 one year ago
Best ResponseYou've already chosen the best response.2no problem so you can do: \[\frac{5+10}{5}=\frac{5(1+2)}{5}=1+2=3 \text{ but you can't do } \frac{\cancel{5}+10}{\cancel{5}}=1+10=11 \\ \text{ and yes another way \to do } \frac{5+10}{5} \text{ is by adding the numerator first } \frac{15}{5}=3\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ahhhhh i see always have to simplify everything on top in the numerator you can't just do 1 and not do the rest... i see now lol thank youuuuuu

freckles
 one year ago
Best ResponseYou've already chosen the best response.2right everything on top has to be divided by the thing on bottom

freckles
 one year ago
Best ResponseYou've already chosen the best response.2like we can't do part of the top divided by the bottom and just ignore the rest being dividing by the bottom

freckles
 one year ago
Best ResponseYou've already chosen the best response.2another way to look at (5+10)/5 is looking at it as 5/5+10/5 separating the fraction first doing the divisions 1+2 then doing the additions

freckles
 one year ago
Best ResponseYou've already chosen the best response.2anyways I think you really do get it now (I think :p) so I will leave you alone now :p
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