All Answers 358150 Answers

Let      log_x3  =  a     --->        x^a  =  3                           (equation 1)

Also:  log₈₁x  =  a     --->     81^a  =  x                            (equation 2)

Substitue equation 2 into equation 1:  (81^a)^a  =  3     --->     81^{a^2}  =  3

Since  81  =  3⁴     --->     (3⁴)^{a^2}  =  3     --->     3^{4a^2}  =  3¹  

                                                                --->        4a²  =  1

                                                                --->          a²  =  ¼

                                                                --->           a  =  ½

Using equation 1:  x^a  =  3     --->     x^½  =  3     --->     x  =  9

https://mbmt.mbhs.edu/

The square root of H lies inbetween (10/3)² and (24/5)²

We first fine 10/3 squared and 24/5 squared. 

10/3 squared is 11.111111111111...

and 24/5 squared is 23.04

Find the amount of integers inbetween. 

12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23

 Count them up and there you have it.

Yay!

Yeah.

My team already solved the problem yesterday.

This person should be banned!!!!!!!! He keeps asking questions from the mongomery blair math tournament online.

Then use the World Wide Web.

Use wikipedia.

History.org.

Therre are so many resources out there!

Don't say you don't have internet.

Is there a picture missing?

I can't help you until you show me where angles A and B are.

They were all kings of different empires.

This is a math site, not a history site, take your history questions to Brainly.

The number of possible multisets is 108.

I would have given a wow react if this was Facebook. 
Few hours ago I answered this person's question, with a full solution.

But anyways thank you for pointing out this person was asking questions that is in a competition. 

This helps to combat cheating. 

After graphing it, the answer is obviously 0.

Graph:
https://www.desmos.com/calculator/s9o19kw1ur

a)

we first get 20/2=10 so we have 20^2pi/4-10^2pi/2=400pi/4-100pi/2=100pi-50pi=50pi

b)

we get 50pi/400=pi/x so we get that it is pi/80 because we divide both sides by 50 and simplify it

Stop cheating, MBMT competitor. 

so it is basically \(\sqrt[3]{8}\) and that equals 2 sinc 2*2*2=8 

your welcome

B. (3x)/(x−5)

since we simplify it into (3x(x−2))/((x−2)(x−5)) by dividing (x-2) and then multiplying back, then we divde both sides by (x-2) to get thee answer B. (3x)/(x−5).

Yeah, this is for the MBMT tournament.

If anyone is confused with my solution feel free to ask me how I solved it

Are you sure? I'm pretty sure that is correct? did I make a mistake?

so like what maxwong said we have a^2+b^2+c^2=71 and a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - (ab + bc +ca))  plug that in to get 11*(71-25) and that equals 506.

_{^{Credts: Maxwong for hints}}

Hmmmm... how do you know this? Where is Blair, I never heard of this. If you can let me know that would be great. I would like to compete next year. Also, if its online I owuld suspect rules say that you may use nothing but your own brain... shame on this competitior