anonymous
  • anonymous
rationalize denominator : please help 1.) 4/ sqrt. 7p 2.) srqt.98/x 3.) 9/3 sqrt.y 4.) sqrt.7 / sqrt.5 +7
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
1.) \[4/ \sqrt{7p}\] 2.) \[\sqrt{98}/x\] 3.) \[9/ 3\sqrt{y}\] 4.) \[\sqrt{7}/ \sqrt{5} + 7\]
anonymous
  • anonymous
please hlep , i dont understand them
dumbcow
  • dumbcow
ok the idea is to get the sqrt out of the denominator For 1) multiply by sqrt(7p)/sqrt(7p)

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dumbcow
  • dumbcow
sqrt(x) * sqrt(x) = x
anonymous
  • anonymous
so it would be p ?
dumbcow
  • dumbcow
7p everything within the radical sorry imagine if i said x = 7p ->4*sqrt(7p)/7p that is the rationalised form for the first one
dumbcow
  • dumbcow
For number 4 you have to use a conjugate which is basically the denominator except you flip the sign ->sqrt(5)-7 multiply this on top and bottom by distributing (using FOIL) and the sqrts in denominator will go away
dumbcow
  • dumbcow
number 2 is already rationalised but you can simplify sqrt(98) by factoring out a perfect square number 3 can be done same way as num 1, dont forget to reduce hope this helps
anonymous
  • anonymous
so how would it be done ?
anonymous
  • anonymous
and thank you for helping , im just really bad in math
dumbcow
  • dumbcow
which part are you stuck on
dumbcow
  • dumbcow
figure it out??
anonymous
  • anonymous
all of it , especially foil
dumbcow
  • dumbcow
ok lets look at num 2 since there is no sqrt in denominator, we don't have to do anything there but we can simplify sqrt(98) Factors of 98 = 2*49 =2*7*7 rewrite sqrt(98) as sqrt(2)*sqrt(7)*sqrt(7) remember sqrt(7)*sqrt(7) = 7 leaving 7sqrt(2)
anonymous
  • anonymous
7sqrt. 2x
anonymous
  • anonymous
?
anonymous
  • anonymous
for number one i got : 4sqrt.7p/7p ? correct ?
dumbcow
  • dumbcow
7sqrt(2)/x sorry no the x doesnt change it stays on the bottom
dumbcow
  • dumbcow
correct
dumbcow
  • dumbcow
for num 3 we want to get rid of sqrt(y) by multiplying by sqrt(y) on top and bottom ->9/3sqrt(y) * sqrt(y)/sqrt(y) = 9sqrt(y)/3y Now reduce what does 9/3 reduce to?
anonymous
  • anonymous
or would 7p cancel itself ?
anonymous
  • anonymous
to 3
dumbcow
  • dumbcow
correct ohh no the 7p wouldnt cancel because the one on top is inside the sqrt, that is important
dumbcow
  • dumbcow
answer for 3) is 3sqrt(y)/y
anonymous
  • anonymous
oh no your answer for 3 is wrong , it doesnt come out in my exam as an option
dumbcow
  • dumbcow
hmm maybe the sqrt is in the numerator try 3sqrt(y)
anonymous
  • anonymous
there is no 3 , only 9 sqrt
dumbcow
  • dumbcow
hmm maybe the sqrt is in the numerator try 3sqrt(y)
dumbcow
  • dumbcow
interesting, what are the possible answers they give
anonymous
  • anonymous
number 3 : A) 9^3 sqrt.y^2 /y b.) 9y c.) 9^3 sqrt. y^2 d.) 9^3 sqrt. y /y
dumbcow
  • dumbcow
and your sure the question was 9/(3sqrt(y))
anonymous
  • anonymous
yes
anonymous
  • anonymous
it better view higher above ^^^^ with the equation button
anonymous
  • anonymous
all the way up^^^^^^^^^^^^^^^^^^^^^^
dumbcow
  • dumbcow
well in that case the answer is 3sqrt(y)/y i'd bet on d) , weird you might have to talk to teacher on this one
anonymous
  • anonymous
and for the rest of the problems ?
dumbcow
  • dumbcow
well what about the first 2 are those answers available :)
anonymous
  • anonymous
do you know how to solve this though ?
dumbcow
  • dumbcow
rationalise fractions with sq roots? yes i know how to solve those but maybe they're looking for something else look at prev examples and see if it lines up with what i've shown
anonymous
  • anonymous
correction : \[9 / ^{3} \sqrt{y}\]
anonymous
  • anonymous
so it changes everything ?
dumbcow
  • dumbcow
little bit the answer is A think of it now as fractional exponents cube root of y = y^(1/3) to cancel that we multiply top and bottom by y^2/3 ->y^1/3 * y^2/3 = y (add the exponents) leaving (9*y^2/3)/y
anonymous
  • anonymous
you mean : 9^3 sqrt.y^2 / y ?
dumbcow
  • dumbcow
oh y^2/3 can be written as cube root of y^2
anonymous
  • anonymous
its the same than ?
dumbcow
  • dumbcow
yes
anonymous
  • anonymous
thank youuu ! its so complicated to me. and the rest ?
dumbcow
  • dumbcow
oh yes FOIL is just way to remember how to distribute. what is result of (x+2)*(x-3)? _>distribute the x to both the x and -3 ->x*x + x*-3 = x^2 - 3x ->then distribute the 2 to the x and -3 ->2*x + 2*-3 = 2x -6 put it together ->x^2 -3x +2x -6 = x^2 -x -6 make sense?
anonymous
  • anonymous
is there an easier way ?
anonymous
  • anonymous
i have to use (x+2)*(x-3) in the equation to cancel out sqrt ?
dumbcow
  • dumbcow
no that was just an example and not really you still have to do the multiplications For this problem we would be multiplying (sqrt(5)+7) * (sqrt(5) - 7)
dumbcow
  • dumbcow
sorry i gotta go the answer is (7sqrt(7) - sqrt(35))/44 try to figure it out on your own by multiplying what i put above for denominator then multiply sqrt(7)*(sqrt(5)-7) for numerator Then at the end you have to distribute a negative by flipping the signs good luck

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