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There is no value of r that makes this true

See the graph here (in  x)  of  the function   :     floor (r)  + r  - 15.5 - floor (r/4)

https://www.desmos.com/calculator/sb9nm8plvj

If we set this function to 0,  the graph never intercepts the  x  axis

So....no solution

what is the sum of all three-digit odd integer palindromes of the form 1A1?  

If I understand the question correctly, this would be:  

     111 

     131 

     151 

     171 

+   191 

———— 

     755 

thanksssss 

Continuing where the first answerer left off...

\({f(2a)}\ =\ 2^{(2a+1)}\\~\\ \phantom{f(2a)}\ =\ \frac{2}{2}\cdot2^{(2a+1)}\\~\\ \phantom{f(2a)}\ =\ \frac{2\ \cdot\ 2^{(2a+1)}}{2}\\~\\ \phantom{f(2a)}\ =\ \frac{2^1\ \cdot\ 2^{(2a+1)}}{2}\\~\\ \phantom{f(2a)}\ =\ \frac{2^{(2a+1)+1}}{2}\\~\\ \phantom{f(2a)}\ =\ \frac{2^{(2a+2)}}{2}\\~\\ \phantom{f(2a)}\ =\ \frac{2^{2(a+1)}}{2}\\~\\ \phantom{f(2a)}\ =\ \frac{(2^{(a+1)})^2}{2}\\~\\ \phantom{f(2a)}\ =\ \frac{(f(a))^2}{2}\)

Hi Guest!
You simply replace "a" with "2a" 

Given: \(f(a)=2^{(a+1)}\) (*)

Then,

\(f(2a)=2^{(2a+1)}\)

For example if you want: \(f(1000a)\), all you do is replace "a" with 2a" in the given (*):

\(f(1000a)=2^{(1000a+1)}\)

grs75.

I owe you an appology.

The question that you are now asking about, that Hectictar answered, was not your question.

So there was no reason why you ever needed to respond to it.  

Maybe you only just saw it. 

I had it confused with this one that Max answered that you did not respond to.

Determine where f is continuous:

Hello Guest!

1. f(x) = sin-1 (2x)\(f(x)=sin^{-1}(2x)\)

    f(x) is continuous if:   \(\color{blue}x\in \mathbb R\ |\ x\notin \{n\cdot \pi /2\}.\)

2. \(f(x) = \sqrt{arctan\ x} / (x^2 - 9)\)

   \(\sqrt{arctan\ x}/(x^2-9)\)  is continuous if: \(\color{blue}x\in \mathbb R\ |\ 0\le x< \infty\ and\ x\ne 3.\)

  !

It is 124

Thanks Grs75.

Your response is appreciated.

It seems my part A and your given answer are the same anyway.

That is a good start :)

Sorry, I most likely missed his response. Additionally, this is from more than a year ago- I am unable to change anything.

I'm sorry. I was expecting to receive an email notification but did not get one, and I am not on this website 24/7. I did not intend to come off as unappreciative. 

As for the correct answers, the answer to part A is 20. A pair of white chairs must be separated by one red chair. The red chairs can be arranged like _ R _ R _ R _ R _ R _ with 6 spaces to put the white chairs. Then 6 choose 3 = 20. Part B is not so clear to me, so I'm not sure of the correct answer.

OK....good deal   !!!!

Hey CPhill I was looking for AB^2, AC^2 and BC^2

Thank you very much for the answer, I really appreciate it. 

I think I can finish it from here!!!

C

4                      5

A           3                       B

cos B =  6/10  =  3/5   →   sin C =  3/5

sin B =  8/10  =    4/5   →  cos C  =  4/5

tan C  =      sin C  / cos C    =   (3/5) / (4/5) =   3 /  4

A = ( 4cos[ pi/4 ]   , 4 sin [ pi /4 ])     =  (4cos 45° , 4sin 45°)  =      (4 / sqrt( 2)  , 4/sqrt ( 2 )  )

B =  (3 cos [ 5pi/12 ],  3sin [5pi/12 ] )  =  (3cos 75° , 3sin 75°)  = 

(  3 [  sqrt (3) - 1 ] / [ sqrt (8)]  ,  3 [ sqrt (3) + 1] / [sqrt (8)]   )

C = ( 2, pi/12)  =  (2 cos 15° , 2sin 15°) =   (2 sin 75° , 2cos 75°) =  

( 2 [sqrt (3) + 1]/ [sqrt (8)]  ,  2[sqrt (3) -1 ]  / {sqrt (8)  ] 

Vinculum ...   Do you want   AB^2  , AC^2  and BC^2  ??

Or do you mean  (AB)^2 , (AC)^2  and  (BC)^2   ??

Also....these are real points......there shouldn't be any  "i"    involved

1 - Minimum number if integers needed so that at least 2 sum up to 949 is calculated as follows:
1,500 - ceil(949/2) + 2 ==1,027 integers.

2 - Minimum number of integers needed so that at least 2 of them will have a difference of 149 is calculated as follows:
a - If the integer of (1,500/149) mod 2==0[i.e. EVEN], then:
[Ceil(1,500/149) / 2 * 149] + [1,500 mod 149] + 1==756 integers [which is the answer to this particular question]

b - If the integer of (1,500 / 149) mod 2 ==1[i.e., ODD such as (1,500/165), then:
Int[(1,500 / 165) + 1] / 2 * 165 + 1==826 integers [for this hypothetical example]

Note that  ∛16 =   ∛2^4  =  2 ∛2

So....the denominator just becomes

∛2 + 2∛2 =   3∛2

So    we  have

  2

___    multiply num / den   by     2^(2/3)  = ∛(2^2) =  ∛4

3∛2

2 * ∛ 4                2 * ∛4          2 * ∛4           ∛4             4^(1/3) /  3   

_______  =     _______  =  _______  =   _____ =   

3∛2 * ∛4            3 * ∛8           3 * 2              3

A +  B  =   4 + 3   =  7

x cannot be negative.....so......the domain  is  [0 , inf )

The pentagon will be composed of 5 congruent isosceles triangles

Their equal sides  = 12   and  the  included angle between  these sides =  72°

The  area of one of these =   (1/2) (12^2) sin (72°)  =   72sin 72°

So the area is ....5  (72)sin (72°)  =   360 sin (72°)   ≈    342.38   cm^2

x - y  =   36         (1)

4y - 3   =  x       (2)

Sub (2)  into (1)

(4y - 3 )  - y    =  36

3y - 3  =  36

3y  = 36 + 3

3y = 39

y =  39 / 3   =   13  =  smaller

And   4(13) - 3  =  x  =  52 - 3  =  49  = larger

A)   We are using the property     ^c √  (a^b)    =   a^(b/c)....so...

  ⁴ √ x^3 =    x^(3/4)

B)  We are using the property that   1 / a^-n  =   a^n .....so.....

1 / x^(-1)   =  x^1   

C)  We are using the property  that   a^m * a^n  =  a^(m+n)

¹⁰ √ (x^5  * x^4  * x^2 )  =    

¹⁰ √ (x^11)  =

x^(11/10)

D)  x^(1/3) * x^(1/3) * x^(1/3)   =

      x^(1/3 + 1/3 + 1/3)  =

     x^1