anonymous
  • anonymous
Which one is greater? =)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\frac{ 1 }{ \sqrt{1} }+\sqrt{\frac{ 1 }{ 3 }}\] or \[\frac{ 1 }{ \sqrt{1} }+\sqrt{\frac{ 1 }{ 4 }}\]
anonymous
  • anonymous
i got the second one as bigger
ganeshie8
  • ganeshie8
which one is big, 1/3 or 1/4 ?

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More answers

ganeshie8
  • ganeshie8
would you like to have 1/3 of cheesecake or 1/4 of cheesecake ?
anonymous
  • anonymous
im on a diet so 1/4
anonymous
  • anonymous
but 1/3 is bigger lol
ganeshie8
  • ganeshie8
so...
anonymous
  • anonymous
wait but can u take a look at what i did
anonymous
  • anonymous
|dw:1437674327047:dw|
ganeshie8
  • ganeshie8
we can't cancel like that \[\dfrac{a+b}{b}~~\ne~~a\]
anonymous
  • anonymous
but doesnt it become a 1?
ganeshie8
  • ganeshie8
nope
anonymous
  • anonymous
what does it become then?
ganeshie8
  • ganeshie8
|dw:1437674587867:dw|
anonymous
  • anonymous
then whats the proper way to do it?
ganeshie8
  • ganeshie8
you 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
  • anonymous
but can you solve it out? I need to know why....
anonymous
  • anonymous
like how do the solutions compare?
ganeshie8
  • ganeshie8
im not getting any ideas, not seeing any neat/standard way for problems like these @freckles
ganeshie8
  • ganeshie8
@dan815
anonymous
  • anonymous
ok can u help me with another problem then?
anonymous
  • anonymous
it's an easy question but i don't know why this website is giving a wrong answer....
anonymous
  • anonymous
If \[(x+3)^{2}=25\] which of the following could be the value of x? a. -8 b. -5 c. -2 d. 5 e. 8
anonymous
  • anonymous
i thought it was this at first |dw:1437675183398:dw|
anonymous
  • anonymous
then i thought it was this : |dw:1437675254689:dw|
anonymous
  • anonymous
but the correct answer is 8... and i don't know why
freckles
  • freckles
@ganeshie8 oh how to compare those two numbers?
anonymous
  • anonymous
Hi 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
  • anonymous
I gues i did it the wrong way but fail to understand why it's wrong
freckles
  • freckles
first 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
  • anonymous
hmmmm okay..
anonymous
  • anonymous
that still wasn't solved tho =(
freckles
  • freckles
what do you mean ?
freckles
  • freckles
i guess I'm asking what are you looking for?
anonymous
  • anonymous
like the problem solved out
freckles
  • freckles
to me I showed why sqrt(1/4)3 which means 1/4<1/3 taking square root of both sides the direction of inequality still holds
anonymous
  • anonymous
hmmm okay i gotcha can you help me with the second problem please?
anonymous
  • anonymous
i just really fon't understand why it's 8 instead of 2 or -8
anonymous
  • anonymous
don't*
freckles
  • freckles
|dw:1437675933396:dw|
freckles
  • freckles
so you are right it is 2 or -8
anonymous
  • anonymous
wait what? is this for all square root problems? it's either a 5 or -5?
anonymous
  • anonymous
oh okay but why is it positive 8 then?
freckles
  • freckles
\[x^2=a \text{ if } a \text{ is positive then yes } x=\sqrt{a} \text{ or } x=-\sqrt{a} \\ \text{ since both } (\sqrt{a})^2=a \text{ and } (-\sqrt{a})^2=a\]
freckles
  • freckles
I 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
  • anonymous
so you're saying it can be 2 and -2 and also 8 and -8?
freckles
  • freckles
no*
freckles
  • freckles
\[(x+3)^2=25 \text{ gives us } x+3=5 \text{ or } x+3=-5 \] is what I'm saying
freckles
  • freckles
\[\text{ which gives } x=5-3 \text{ or } x=-5-3 \]
anonymous
  • anonymous
yea
anonymous
  • anonymous
so positive 8 is wrong?
freckles
  • freckles
yeah 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
  • anonymous
ahhhhh okay i gotcha ^_^ thank youuuuuuuuuuuuuu!!~~~
freckles
  • freckles
can we go back to the previous problem @yomamabf were you expecting a certain method on that one?
anonymous
  • anonymous
Well 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
  • anonymous
|dw:1437676640564:dw|
anonymous
  • anonymous
like i don't understand why thats wrong
freckles
  • freckles
these look different then the other two values you gave
freckles
  • freckles
\[\frac{1}{\sqrt{1}}+\sqrt{\frac{1}{4}} \text{ or } \frac{1}{\sqrt{1}}+\sqrt{\frac{1}{3}} ?\] are these not the two values you were comparing ?
anonymous
  • anonymous
|dw:1437676741343:dw|
anonymous
  • anonymous
yea those are but i just solved it is why they look different
freckles
  • freckles
\[\frac{1}{\sqrt{1}}+\frac{\sqrt{1}}{\sqrt{3}} \\ \] 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
  • freckles
you can "cancel out" common factors in division
freckles
  • freckles
not common terms
anonymous
  • anonymous
oh okay so u can cancel out if you divide but cant reduce in this situation because we're not dividing?
freckles
  • freckles
fraction bar is division
anonymous
  • anonymous
so 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
  • freckles
\[\text{ say you have this } \\ \frac{\sqrt{3}+2 \sqrt{3}}{\sqrt{3}} \\ \text{ the \top and bottom share a common factor } \sqrt{3} \\ \ \text{ that is you can factor out a } \sqrt{3} \text{ on \top } \\ \frac{\sqrt{3}(1+2)}{\sqrt{3}} \\ \text{ notice top and bottom have } \sqrt{3} \text{ factor \in common } \\ \text{ we know } \frac{\sqrt{3}}{\sqrt{3}} =1 \text{ so we can do } \\ \frac{\cancel{\sqrt{3}}(2+1)}{\cancel{3}}=\frac{2+1}{1}=2+1=3\]
anonymous
  • anonymous
Like for this one |dw:1437677209526:dw|
freckles
  • freckles
basically all terms have the have the same factor for example: \[\frac{6x^3+5x^2+x}{2x^2-x}\] 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 \[\frac{6x^2+5x+1}{2x-1}\]
anonymous
  • anonymous
yea i understand that but for roots its a bit different right?
freckles
  • freckles
nope
freckles
  • freckles
\[\frac{ 6 (\sqrt{3})^3+5 (\sqrt{3})^2+\sqrt{3}}{2 (\sqrt{3})^2-\sqrt{3}} \text{ divide top and bottom by } \sqrt{3} \\ \text{ divide each term in the top by } \sqrt{3} \\ \text{ divide each term in the bottom by } \sqrt{3} \\ \frac{6 (\sqrt{3}))^2+5 \sqrt{3}+1}{2 \sqrt{3}-1}\]
anonymous
  • anonymous
hey for the problem u put on top aren't u supposed to solve it like this? |dw:1437677469609:dw|
anonymous
  • anonymous
|dw:1437677555931:dw|
freckles
  • freckles
\[\frac{\sqrt{3}+1}{\sqrt{3}} \\ \text{ if you did \choose to try what you did earlier } \\ \text{ you would also need to divide the } 1 \text{ by } \sqrt{3} \\ \text{ that is you would have } \\ \frac{\frac{\sqrt{3}}{\sqrt{3}}+\frac{1}{\sqrt{3}}}{ \frac{\sqrt{3}}{\sqrt{3}}} =\frac{1+\frac{1}{\sqrt{3}}}{1}=1+\frac{1}{\sqrt{3}} \text{ but this is what you had to \begin with } \\ \text{ so you were working backwards to what you started with }\]
freckles
  • freckles
you could but another way is the way I did
freckles
  • freckles
and the way I did it was to give you an example of when you can cancel
anonymous
  • anonymous
if it involves addition and subtraction don't reduce it but if it involves multiplication ad division then reduce it
anonymous
  • anonymous
oh wait nevermind HAHAHAHAHHAHA OHHHHHHHHHHHHHHHHHHH IM SO STUPID OKAY I GOT IT
anonymous
  • anonymous
SORRY hahahaha okay i got it lmaoooooo
anonymous
  • anonymous
i see what i did wrong this is what it's supposed to be |dw:1437677816464:dw|
anonymous
  • anonymous
ahhhh i see lolol
anonymous
  • anonymous
thank you so much for breaking everything down you're the best <33333
freckles
  • freckles
no 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
  • anonymous
ahhhhh 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
  • freckles
right everything on top has to be divided by the thing on bottom
freckles
  • freckles
like we can't do part of the top divided by the bottom and just ignore the rest being dividing by the bottom
anonymous
  • anonymous
lol gotcha <3
freckles
  • freckles
another 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
anonymous
  • anonymous
kk
freckles
  • freckles
anyways I think you really do get it now (I think :p) so I will leave you alone now :p
anonymous
  • anonymous
thank u again =)
freckles
  • freckles
np

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