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Are a bunch of posters here on the same home-schooling program?  Because I am seeing a LOT of multiple posts of the EXACT same questions from DIFFERENT posters for the last few days....and they are posted with the EXACT same fonts !

Hmmmmm......

\(\frac{f}{r^2}+a=n+k\\ \frac{f}{r^2}=n+k-a\\ f=r^2(n+k-a)\\ \frac{f}{n+k-a}=r^2\\ r=\pm\sqrt{\frac{f}{n+k-a}}\)

Your answer is correct!

√ 24x  * √ 18y  *  √ 15y  

2√ 6x  * 3√ 2y * √ 15y  

6 √  [ 6x * 2y * 15y ]

6 √  [ 180 * x * y^2 ]  

6 √  [ 36 * 5 * x * y^2 ]

36y√ [ 5x ]  ⇒  Correct  !!!

YOU GOT IT!.

Correct......you're doing well  !!!

A)   top part  = cone   =  pi * r^2 * h  / 3

       bottom part  = cylinder  =   pi * r^2 * h

B) top part volume  =  {  the radius  = diameter / 2}   =  3.1416 * (27/2)^2 * 9 / 3  ≈

1718 ft^3

C) bottom part volume  =  3.1416 * (27/2)^2 * 21 ≈ 12024  ft^3

Total volume  =   ( 1718 + 12024)  ft^3   =   13742 ft^3 

1) lne  = 1

2) lne²  = 2 ln e    =  2 (1)  = 2

3) lne¹⁰ = 10 ln e   =  10 (1)  = 10

4) 10 lne   =  10

5) ln1   =  0

6) lne/4  = 1/ 4

7) lne²/2    =  2 ln e  / 2   =  ln e

8) lne⁸³  =  83 ln e    =  83 (1)  =  83

I do not know what  'X' and 'Box' are...if they are a form of factoring....SOMEone taught you those....right?   Wasn't it a teacher???

Correct....!!!

No doubt....A math question you're allowed to answer only if you have CIA clearance

Unfortunately, I am confronted with this message when I try the provided link.

"You have requested a top secret page for which you have not received security clearance. Or maybe you have clearance, but you are not logged in. Click here to log in."

Correct...( to the nearest degree...)

Arc length  =      2pi (radius) (x / 360) 

Where x is the angle(in degrees) we're looking for

So  we have

15 pi  =  2pi (60) (x /360)     divide pi out and simplify

15  =  2 (60) * x /360      multiply both sides by 360

5400  = 120 x     divde both sides by 120

45°  =  x

1) 4log₃2 - 2log₃x = 1

Note that  1   =  log₃ 3  .....so we can write

4 log₃ 2 - 2log₃x  = log₃ 3      {  log property :  b log a  =  loga^b  }

log₃ 2^4  - log₃x^2  = log₃3   { log a - log b  =  log (a/b) }   ⇒ 2⁴  = 16

log₃ (16/ x^2)  = log₃ 3

Since  we have equal logs   we can just solve

16/x^2  =  3

x^2  = 16 / 3         take the positive root 

x  =  4 / √ 3   ≈ 2.3094

2) logx² = 2       ....we can write

2 log x  =  2         divide both sides by 2

log x  =  1

This says that    10^1  =  x  =  10

Thank you

Dang.....I hate typing in my answer when I cannot see the question on my screen...I got the turn 3 backwards......so here it goes a gain.....

He has two option   roll flip flip    or  roll roll flip

1st roll has a 2/6 chance of being a 1 or a 2    2/6 = 1/3

  then he flips a coin and has a 1/2 chance it is a HEADS   (so he would flip coin again turn 3)

He has a 4/6 chance of NOT rolling a 1 or a 2 ...... which means turn 2 would be a roll again

  then he has a 2 out of 6  (1/3 chance of rolling a 1 or a two)  which makes turn 3 a coin flip

1/3 * 1/2    +   4/6 * 1/3 =  1/6 + 4/18 = 7/18 chance of coin flip on turn 3

This is my first answer that was a  WRONG answer:

Well.....no one else has posted an answer to this....so I will make my usual probability-guess at it!

He has 2 out of 6 chance of rolling a 1 or a 2   on the die......  2/6 = 1/3

    then when he flips the coin, he has a 1/2 chance of tails   (which would make turn 3 a roll of the die)

He has a 4/6 chance of NOT rolling a 1 or a 2  (which means he ROLLS again)

   His second turn will  be a roll.... and he has a 4/6 chance this will NOT be a 1 or a 2...

        which would make turn 3 a roll of the die

so

1/3 x 1/2    +   4/6 * 4/6  = 1/6 + 16/36 = 22/36 = 11/18 chance turn 3 will be a roll (??)

y^2  =  3x + 10

Subtract 10  from both sides

3x  =  y^2 - 10      divide both sides by 3

x  =  (1/3)y^2  - 10/3

This is a  "horizontal" parabola that opens to the right....

I'm sure there is....but....I'm not a good enough geometer to figure it out....

This  is a circle centered at the origin  with a radius  of  9

Thanks, but is there a way to not use trig?

Total number of gallons in the pool

Total number of minutes filling the pool

Find the center of the circle.....call this  " O"

From O.....draw  bisector OG  to chord AB   and bisector OH  to chord EF 

But, by Euclid,  a radial line bisecting a chord does so at right angles

And since AB  = EF = CD.....then equal chords are equal distances from the center of a circle

Now....connect  O  and  P

So.....OG  = OH,  OP  = OP   and angles OGP  and OHP  are right

So, by HL,  have two  cogruent  right triangles formed, ΔOGP  and ΔOHP

But  angle GPO   =  angle HPO....so  angle P  is bisected

By similar reasoning, we can draw bisector  OI  to chord  CD  and  also connect  QO

This will form  two congruent right triangles - Δ OGQ and ΔOIQ

And angle GQO  =  angle IQO, so Q is bisected

And since PO  and QO are angle bisectors of triangle  QPR  that  meet  at O....then ,by definition, the center of the circle is the incenter of triangle PQR 

Just multiply the binomial and trinomial together. 

\((x-3)(2x^2-x+3)\\ 2x^3-x^2+3x-6x^2+3x-9\\ 2x^3-7x^2+6x-9\)