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I don't understand your logic 

\(\text{Suppose you want the numbers of 3 consecutive tiles to sum to $N$}\\ \text{Further let's assume that the set you refer to has $M$ numbers in it}\\ \text{We want the number of 3 element partitions of $N$ from the elements of the set}\\ \text{This is equivalent to sorting $N$ balls into $M$ bins}\\ \text{The number of ways this can be done is given by the stars and bars problem (google) and is}\\ \dbinom{N+M-1}{M-1}\\ \text{Tiles can additionally be colored red or blue. This gives an additional factor of $2^L,\\$ where $L$ is the length of the row of tiles.}\\ \text{So if your row is length $L$ the number of arrangements is }\\ 2^L \dbinom{N+M-1}{M-1}\)

\(v_1 = (1-0.2)v_0 = 0.8v_0\\ v_2 = (1+0.3)v_1 = 1.3(0.8)v_0 =1.04v_0\\ \dfrac{v_2}{v_0} = 1.04 = (1+0.04)\\ \text{Overall there was a 4% increase in the stock value}\)

True

all but one of the prime numbers are odd.

True.

2 is prime.

True.

2 is the only even prime.

The statement is false.

(x-2)^(2/3) =  36   raise  both sides to the  3/2   power

[ (x - 2) ^(2/3) ]^(3/2)   =  36 ^(3/2)

(x - 2)^(2/3 * 3/2)  =  ( √ [ 36] )^3

(x - 2) ^1   =  (6)^3

x - 2 =  216     add 2 to  both sides

x   = 218

Check

(218 - 2)^(2/3)   =36   ???

(216)^(2/3)  = 36  ???

[ (216)^(1/3) ] ^2  = 36  ???

[ ∛216] ^2  = 36  ???

[6]^2   = 36        true   !!!!

At the x value of n, n^2+2 has to be equal to 2n+5 for the function to be continuous at x=n.

\(n^2+2=2n+5\)

\(n^2-2n-3=0\)

\((n-3)(n+1)=0\)

\(n=3, n=-1\)

The sum of all values of n is 3-1=2.

Yes....Melody's answer is  correct.....I didn't account for the  fact that we want the ratio of even palindormes 4-digit palindromes to ALL 4-digit  integers... 

Thanks, Melody!!!!

I think you forgot to type in what the numbers sum to.

Actually Juriemagic, you can just post your image in imgur or imbb or something like that. Imgur and ImgBB are websites where you can upload any image saved to your PC. Then copy the direct link and paste it in when you are posting an image.

1)   10920/803<=r<10940/803

1 -  r = 7.43

Floor[100 * 7.43] = 743.

2)

Solve for x:
2008 abs(x^2 - x + 1/2008) = 1

Divide both sides by 2008:
abs(x^2 - x + 1/2008) = 1/2008

Split the equation into two possible cases:
x^2 - x + 1/2008 = 1/2008 or x^2 - x + 1/2008 = -1/2008

Subtract 1/2008 from both sides:
x^2 - x = 0 or x^2 - x + 1/2008 = -1/2008

Factor x from the left hand side:
x (x - 1) = 0 or x^2 - x + 1/2008 = -1/2008

Split into two equations:
x - 1 = 0 or x = 0 or x^2 - x + 1/2008 = -1/2008

Add 1 to both sides:
x = 1 or x = 0 or x^2 - x + 1/2008 = -1/2008

Subtract 1/2008 from both sides:
x = 1 or x = 0 or x^2 - x = -1/1004

Add 1/4 to both sides:
x = 1 or x = 0 or x^2 - x + 1/4 = 125/502

Write the left hand side as a square:
x = 1 or x = 0 or (x - 1/2)^2 = 125/502

Take the square root of both sides:
x = 1 or x = 0 or x - 1/2 = 5 sqrt(5/502) or x - 1/2 = -5 sqrt(5/502)

Add 1/2 to both sides:
x = 1 or x = 0 or x = 1/2 + 5 sqrt(5/502) or x - 1/2 = -5 sqrt(5/502)
Add 1/2 to both sides:
x = 1 or x = 0 or x = 1/2 + 5 sqrt(5/502) or x = 1/2 - 5 sqrt(5/502)
Sum of squares= 1^2 + [1/2 + 5 sqrt(5/502)]^2 + [1/2 - 5 sqrt(5/502)]^2 =1003 / 502

Suppose a is directly proportional to b, but inversely proportional to c. If a=2 when b=5 and c=9, then what is c when b=3?

\(\color{BrickRed}a:b=2:5\\ \color{BrickRed}a:\frac{1}{c}=2:\frac{1}{9}\\ a=\frac{2b}{5}\\ \frac{2}{c}=\frac{a}{9}\\ c=\frac{18}{a}\)

b=3

\(a=\frac{2b}{5}\\ a=\frac{6}{5}\\ c=\frac{18}{a}\\ c=\frac{18\cdot 5}{6}\\ \color{blue}c=15\)

Thanks Melody! I have "looked up".

  !

CPhill has found the probability that it is an even palidrome given that it is a palindrome.

I do think think this is quite what the question is asking.

I think it wants the probability that it is a 4 digit even palidrone given that it is a 4 digit number.

The number of 4 digit numbers is 9*10*10*10 = 9000

The answer I believe is 

4/900

Asinus, you need to look up what inversely proportional means.

He is right, this is not correct. I have looked through your work however, and I can't see what you did wrong. 

Thanks Guest!!!

Thank you!!!

Thanks all of yyou!!!

Thanks for the help!!!