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log _x 32  = 5/2

Using the change of base rule we have that

log 32               5

_____     =      ___           rearrange as

log x                  2

(2/5) log 32  = log x

This says that    10^[ (2/5)log 32 ]  = x  =  4

Angle CKJ is paired wth angle IKD

how did you get how many odd numbers there are?

What about angles such as CKJ? you left that out... We are nvr going to finish counting...

😭😭😭😭😭😭😭😭😭😭

What is the minimum surface area that suc---

wat?

Is it just me or Do i not get it...

:( I wish i got it right though- I treid counting and i got alot more that 30 but i stopped- it was 2 messy...

Sorry....I don't remember.....!!!

I do- im still pretty young and growing up seems great so far... im now 11 though! yayyyy

yeah but what was it? my first one was 10.

My first answer was incorrect, tom......maybe someone else has another approach/answer

EP is correct!!!!

BTW..... we can verify that  s = 14  produces a minimum surface area by taking the derivative  of

2s -  5488s^(-2)  =

2  + 10976s^(-3)  

When s = 14    this is  > 0     which means that s = 14  produces a minimum surface area

How many bases $b$ are there such that $663_b$ is prime?   

Let "b" stand for any base.

663 = (6 x b²) + (6 x b) + 3  

the above can be written 6b² + 6b + 3 

you can always factor a 3 out of 6b² + 6b + 3

so, there is no base in which 663 is prime 

.

Yeah is there another answer you got? before rom corrected you last time?

I saw your answer on mellies... I thought it made sense, but now i need to get it right so I can look at the solution.

And basically figure it out...

These angles will have a corresponding pair  

AKE , IKA, GKI, DKG , FKD

EKI, AKG, IKD, GKF, DKB

EKG, AKD IKF, GKB , DKJ

EKD , AKF, IKB, GKJ, DKH

EKF, AKB, IKJ, GKH,DKC

I get  n^2  = 25, as well, tom......

Fewer dollar signs version of the problem:

When \(p(x) = Ax^5 + Bx^3 + Cx + 4\) is divided by \(x - 3,\) the remainder is 11. Find the remainder when \(p(x)\) is divided by \(x+3\).

The dollar signs are ok for now, but when you get to like a huge problem with a fraction polynomial, we really can't understand it... That would not be the most ideal form of the problem...

That may result in a wrong answer...

THX,  EP  and hectictar  !!!!

I agree with EP. The locus would be the four points of intersection of the circle and the blue lines in this graph:

See the following image......

Let  A = (6,30)   B = (0, 0)   C = ( 12,0)   D  = (6,0)    and AD  = 30

Since ED is parallel to AB.....then ang;e EDC  = angle ABC

And angle ACB  = angle ECD

So triangle ABC will be similar to triangle EDC

And  the base of EDC  = DC is 1/2 of base BC....so

The area of triangle EDC  = Area of triangle ABC * (scale factor of EDC to ABC)^2  =

!80 * (1/2)^2  =   180 * (1/4)  = 45 units^2

So....[ ABDE]  = [ ABC ] - [EDC]  =  180 - 45   = 135 units^2

Dividing p(x) by x - 3 implies that

p(3)  =   Remainder  ....so....

A(3)^5  + B(3)^3 + C(3) + 4  =  11  ⇒   243A + 27B +3C + 4  = 11 ⇒ [ 243A + 27B + 3C]  =  7

And p(-3)  = Remainder

A(-3)^5  + B(-3)^3 + C(-3) + 4  = Remainder   ⇒  -243A - 27B - 3C  =  Remainder - 4  ⇒

- [ 243A - 27B + 3C]  = Remainder - 4

- [ 7 ]    = Remainder - 4      add 4 to both sides

-3  = Remainder =  p(-3) 

x  = 2y^2 - 6y  + 3

The x intercept is easy....let y  = 0   and the x intercept is  (3, 0)   = (a, 0)

To find the y intercepts.....let x = 0   and solve for y

2y^2 - 6y +  3 = 0

Use the quadratic formula

y  =     6 ±√[ 6^2 - 4*3 * 2 ]             6  ±√ [ 36 - 24]             6 ±√12

          ________________  =    ______________  =   __________

                 2 *2                                      4                              4

The sum of these two  roots  = b + c   =    2 (6/4)  =  3

So

a + ( b + c)   =  3 + (3)   =   6

Rearrange  as

x^2 - 8x + y^2 + 6y  = -20       complete the square on x and y

x^2 - 8x + 16  + y^2 + 6y + 9  =  -20 + 16  + 9

(x - 4)^2  + ( y + 3)^2  =  5

The center is  ( 4, -3)  =  ( x , y)

So    

x + y   =   4 + (-3)   =    1