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I stopped grinding math a few months ago cuz like im lazy

REET its time to grind math! I still dont think its right!!!!!

We know that P(a) - P(b) is divisible by a - b.  This means if a - b is not a prime number, then P(a) - P(b) is not a prime number.  So by modular arithmetic, we can find a positive integer n such that P(n) is composite.

If P(a_1) and P(a_2) are composite, then P(a_1) - P(a_2) will be divisible by some prime number.  Similarly, if P(a_1) - P(a_2) is divisible by some prime number, and P(a_1) or P(a_2) is composite, then the other one is also composite.

We can then take a_1 = n.  So there exists a positive integer a_2 such that P(a_2) is composite.  But since P(a_2) is composite, we can apply the result above, to find a positive integer a_3 such that P(a_3) is composite.  In this way, we can generate an infinite number of positive integers n such that P(n) is composite.

Lol same here I stay up until like 4 then wake up at like 10

its 11 pm here. I've become semi-nocturnal since my school shutdown :/

When your so bored that you have a compition between two leaderboard people. Lol its like 9 o clock here.

Aww dont be a bully

I'm not CPhill. Our age difference measures in multiple decades XD trololololololol

3) Circle O has diameter AE with AE = 8 cm. Point C is on the circumference of the circle such that AC and CE are congruent. Also, AC is a diameter of semicircle ABC, and CE is a diameter of semicircle CDE, as shown in the figure. In square centimeters, what is the total combined area of the shaded regions?

Believe it or not, the two shaded regions  have  the same combined area  as  that of right triangle ACE

Proof  :

Radius  of semi-circle  CDE  = (1/2)CE

So...area  of   semi-circle CDE  = (1/2)pi [(1/2)CE]*2 =  (pi / 8) CE^2

And similarly.....the area of semi-circle  ABC =  (pi/8)AC^2

Area  of  right triangle  ACE  =  (1/2)(AC)(CE)

Area  of    two  unshaded   regions  between the right triangle and  the circle  =

(1/2)pi[(1/2)AE]^2  - (1/2)(AC)(CE)  =  

(1/2) pi (1/4) [ AE]^2  - (1/2)(AC)(CE)  =    { note that AE^2  = CE^2 + AC^2 }

(1/8) pi [ CE^2 + AC^2 ]  - (1/2)(AC)(CE)

So....the  area  of  the two shaded regions  =

(pi/8) (CE^2 + AC^2]  -  [ (1/8)pi (CE^2 + AC^2 ) - (1/2)(AC)(CE) ]  =

(1/2)(AC)(CE).....which is the same  area as right triangle ACE

And  the area of  right triangle  ACE = (1/2) (CE)(AC) =   (1/2)(8/√2)(8/√2)  = 64 / 4  =  16 cm^2 = area of two  shaded regions

Hmm Im still confused! I think we did it wrong somewhere! Caculator user was CPhill right? :) Lol

lol I just plugged the values into the web 2.0 calculator cuz i'm lazy

Hmm I think you ment to say this?:^{\(( x\cdot(3x+1) \)} Still the question dosent make any sense so we cant really answer it

the slope is incorrect... think you can figure out the correct slope and work off of guest's answer?

I remember learning this in elementry school! To do this you can first right it out: 1/(1/3). Then the way we learend it was to keep the first number change the symbol and flip the last number. So 1/(1/3) would become: 1*3. So your first answer was correct!

Thats a nice one Cphill! But wouldnt you just do the rest mentaly after you find out t = 4? Its kind of a pain typing it out! There is a typo at the end. Dollars isnt spelled Dolars!

Ha Ha Ha! Its okay Cphill! I think you made a mistake on the other question though! I think its a tough one! I dont think its C,A,B,E,D personaly

Oops....just didn't take it far enough....

m  = (-1/2)    b  = 4

So

m +  b  =  

3 + 1/2  = 

7/2

I have a hunch, but it probably makes no sense.

Maybe the use of Mass Points works here.

Set a mass of two of point Q, one for M and one for P.

If B has a mass of one, then point A must have a different mass, and the two masses add to some number.

However, both mass A and mass C add to point P, which has a mass of one.

D has a mass greater than one.

Finally, that means the ACD triangle area is greater than that of ABP triangle.

I'm sure that there is a way better explanation.

Wait Cphill but you want to find m+b!

Note that  

Angle M  = (1/2) its intercepted arc  =  (1/2)(80)   = 40°

Similarly, angle N = (1/2) 35  =  17.5°

Then in triangle MYN  angle MYN   =180 - 40 - 17.5  = 122.5°

But  angle XYZ  is a vertical  angle to angle MYN  so it also measures 122.5°

A  2^(1/2) / 4^(1/6)  =  2^(1/2) / ( 2^2)^(1/6)  = 2^(1/2) / 2^(1/3)  =  2^(1/2  - 1/3)  = 2^(1/6)

B    (128)^(1/28)  = [ 2^(7)]^(1/28)  = 2^(1/4)

C    (1 / 8^(1/5))^2  =  [ (8)^(-1/5)]^2  =  (2^3)^(-1/5)  = 2^(-3/5)

D     Note    4^(-1)  = 1/4       2^(-1)  = 1/2    8^(-1)  = 1/8

So  what  we have inside the radical is just this :

                1/4                        1/4

             _________   =    _______   =   (1/4) * (16/1)   =   4

              1/2  * 1/8                1/16  

And       the sqrt of this =   2

E     Inside  the radical  we have 

2^(1/2) * 4^(-1/4)  =

2^(1/2) * (2^2)^(-1/4)   =

2^(1/2) * 2^(-1/2)  =

2^(1/2 - 1/2) =

2^0  = 

1

And the cube root of this   is still   just 1

So  in order we have

2^(-3/5) ,  2^(1/6)  ,  2^(1/4) , 1 , 2

C , A , B , E , D