Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

14- Find a decimal approximation to the nearest tenth: i) √82 Is there ANY easy way other than: Find the perfect squares that is before and after the number 82 and then add the both numbers and blah..blah..

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
since it is to the "nearest tenth" you won't have many numbers that you need to try. start with 9 since 9*9=81 and increase by one tenth until you get an answer that is closest to 82.
Can you show me?
another possible method might be to use the fact that:\[(a+b)^2=a^2+2ab+b^2\]so set a=9 and you need to find b such that:\[(9+b)^2=82\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

therefore:\[(9+b)^2=81+2*9*b+b^2=81+18b+b^2\]therefore:\[81+18b+b^2=82\]therefore:\[b^2+18b=1\]then try b=0.1, 0.2, ....
Oh cool =) Can you do this method if they asked you " Find a decimal approximation to the nearest hundredth" ? @asnaseer
yes, then you would need to try b=0.01, 0.02, etc. However, this might take too long, so a better strategy in this case might be to use a technique where you keep halving the interval, e.g.: try b=0.50 ---> if this is too low, then try b=0.75 ---> otherwise try b=0.25 that will converge quicker.
have you heard of the Newton-Raphson method?
if you have, then that will converge to the answer faster.
Actually no I havent
no worries - just use one (or more) of the techniques mentioned above to see which one you prefer.
Thank you so much =D
yw :)
Okay, I got a bit stuck using your method.. My method is SOOOOO complicated. Anyway, right you said you try b=0.1,0.2.. like this: b^2+18b=1 0.1^2=18(0.1)=1 ? and if it didnt work you try 0.2 and if didnt work you try 0.3 and so on? O.o
you won't find an exact answer - you are supposed to find the nearest answer to one tenth
Yeah and I think you have to round it at the end right?
ok, let me try and explain. we said we can re-arrange the question to this form:\[b^2+18b=1\]
but my question is you substitute b with 0.1 like that? until you find an answer to the nearest tenth?
so what we need to do is to find what value of b will give an answer that is "closest" to 1. we can re-arrange further to get:\[b^2+18b-1=0\]so now we need to find a value for b that gives the closest answer to zero. so, lets try b=0.0, we get \(b^2+18b-1=-1\) so, lets try b=0.1, we get \(b^2+18b-1=0.81\) so, lets try b=0.2, we get \(b^2+18b-1=2.64\) so the "closest" answer is when b=0.1 therefore, the answer is 9.1 to nearest tenth.
does that make sense?
It makes sense till so the "closest" answer is when b=0.1 and then how did it become 9.1?
because we started this from:\[(9+b)^2=82\]now we have found 'b', answer is '9+b'
remember we said \(9^2=81\) and we are trying to find a number that, when squared equals 82. so that number must be 9 plus a little bit. we called that "little bit" 'b'
This is even more complicated lol but yeah i understood how =) Thank youu!!
Afwan :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question