Can someone please help? http://prntscr.com/afofwh
@FortyTheRapper
Radical is pretty much square root in human terms Let's begin by finding C using the Pythagorean theorem. Are you familiar with that?
Sorta, I'm learning it in class this week.
So, basically it's \[a^2+b^2=c^2\] We take our value of a and square it, we take our value of b and square it, then we add them up and put a square root symbol over that answer
Alright so, 5=2.2360679775 We still add them up even if they are the same number?
Oh, you took the square root of the number 5, but we want to square the number 5 (Multiply it by itself)
Then you put the sqrt over it \[\sqrt(a^2+b^2)\]
So like 2.2360679775 * 2.2360679775 = 5?
We would do for A, let's say \[5^2\]
B is the same number in this case, so we do the same for B
sqrt(5^2 +5^2) so we just this equation?
Right, so what does 5^2 equal?
25
Right, so now we do sqrt(25+25). 25 + 25 = 50 right, so would you agree that \[C^2 = \sqrt(50)\]
Yes, So that's what I write for C?
For this question, yes \[\sqrt{50} \] is in simplified radical form
I mean unsimplified
Alright awesome, now to part b?
Yep To simplify this, we need to know our perfect squares like 1,4,9, etc 1 = 1*1 4 = 2*2 9 = 3*3 and so on
Okay, so we're going to simplify \[\sqrt{50}\]
right?
Correct
In order to simplify it, we have to find 2 numbers that multiply to 50 One of those numbers, though, has to be a perfect square
\[\sqrt{2*5*5}\]
Like that ^
Yes! Right numbers, we just need to separate them \[\sqrt{2}\sqrt{5*5}\] 5*5 = 25, so \[\sqrt{2}{\sqrt{25}}\]
See how that bottom one equals \[\sqrt{50}\]
\[\sqrt[5]{2}\]
Yes, \[5\sqrt{2}\] is in simplified form, good job
Just remember to make it a big 5 and not a little 5
So thats what goes for part B?
Yep, \[5\sqrt{2}\], you found part B
How do we find it to the nearest hundredth now?
I like part C because C stands for calculator lol Let's put \[\sqrt{50}\] in the calculator and see what we get
It gave 7.07, which is in the hundredths, sounds good for a Part C answer!
Awesome! Thank you so much! Do you think you can help with 1 more problem?
Yeah, let's see it
Alright, lemme open a new thread. :)
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