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-sqrt (64)   =   -8

-cube root ( 81)  =  -3 cube root (3) ≈  - 4.326

4throot (25)  ≈  2.236

cube root (27)  =  3

sqrt (21)  ≈ 4.58

sqrt (40)  =  2sqrt (10)  ≈ 6.32

The sum = 3 + (-2) = 1

t^2 - 9t  - 36   +  3t^2   - t  + 10          simplify

4t^2  - 10t  -  26    complete  the square  on  t

4  ( t^2  -  10/4  t   -  26/4)

4 ( t^2   -  5/2 t   -  13/2 )

4 ( t^2  -  5/2 t  +  25/16  - 13/2 - 25/16)

4 [ (t - 5/4)^2 ]  -  26  - 25/4

4 (t - 5/4)^2   -  129/4

This  will  be minimized  when  t  =  5/4 

12 containers

https://web2.0calc.com/questions/angle-bisectors

I will start you off.

I let the 6th one be y

so the 7th is   10-y   [becasue   5+y+10-y=15]

So the 8th is 5

etc

Keep going and you will be able to work out what x is.

Please other answerers, let guest asker do it for him/her self from here.

that is if the 1 keeps repeating forever.

Don't trust guests.

Btw, Guest below is correct because his name is civonamzuk, but he's lazy to log in!!!

In triangle JKL, we have JK = JL = 25 and KL = 30. Find the circumradius.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Semiperimeter   s = 40 

Triangle area     A = 300          

Inradius    r = A / s     r = 7.5

Circumradius   R = (abc) / 4rs     R = 15 ⁵/₈

a)  Yes     17 + 20 = 37

b)  No       116 + 125 ≠ 231

7 1/9

A) No. B) No

BTW u play roblox or what

Ok, that's what I thought the answers were... Thanks so much. You really have been helpful! Thanks!

Ok, Thanks soooo much! 

helpppppp meeee im stucckkkk

No, theres 38.

Remember these definitions:

Theoretical Probability: This is what is expected to happen. 

Experimental Probability: This is what actually happened in an experiment.

(A) - Since a number cube has \(6 \) even sides, the theoretical probability of rolling a side (\(3\) in this case,) is \(\frac{1}{6}\).

(B) - If the number \(3\) came up \(67\) times out of the \(450\) times rolled, then the experimental probability is \(\frac{67}{450}\).

and now its ded with 3 ppl ngl

After some messing around, I got this:

\(\displaystyle \sum_{k=0}^{\operatorname{floor}\left(\frac{n}{2}\right)+1}\operatorname{nCr}\left(n,\ 2k\right)\cdot2^{\left(n-2k\right)}\)

Basically, the floor(n/2)+1 calculates the number of possible twos given n. For example, if n is 3, the number of possible twos is floor(3/2)+1=2 (which is correct, since there could be either 0 or 2 twos)

The nCr(n, 2k) basically chooses the spots for the twos to be in.

Lastly, the 2^(n-2k) fills the rest of the remaining spots with either 3s or 1s. 

It does not look very neat, but I don't really have the effort and determination to reduce it though lol

ok thx! ...sometimes maybe i will let someone else answer and see if ur answer and theirs are the same!

and tertre...

-cube root 81, -square root 64, quad-root 25, cube root 27, square rroot 21, square root 40.

Hi, guest! I'm answering right now! It might take some time. . . 

Yes, I think you are correct because \(\frac{1}{4}\)'s of the letters are a. This isn't as hard as combinatorics or permutations, like when you have to choose multiple objects or find out how many total ways to arrange them, etc. ... 

AnxiousLlama