All Answers 358139 Answers

0 + 5 = 5

5 + 7 = 12

12 + 9 = 21

21+ 11 = 32

32 + 13 = 45.

The pattern hides in the bold numbers :)

Yes I figured that you could already do them :)

It is nice to see you here Max :)

x=1+\dfrac{1}{2+\frac{1}{1+...}}\\ 1+\dfrac{1}{x}=1+\dfrac{1}{1+\frac{1}{2+...}}\\

\(x=1+\dfrac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+...}}}}\\ x-1=\dfrac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+...}}}}\\ x-1=\dfrac{1}{1+\frac{1}{2+(x-1)}}\\ x-1=\dfrac{1}{\frac{2+x-1+1}{2+x-1}}\\ x-1=\dfrac{1}{\frac{2+x}{1+x}}\\ x-1=\dfrac{1+x}{x+2}\\ (x-1)(x+2)=x+1\\ x^2+x-2=x+1\\ x^2=3\\ x=\pm\sqrt3\)

As our guest said, all the signs are positive so the answer is    \(\sqrt3\)

-8*(-3+15)

=-8*(12)

=-96

:D

~The smartest cookie in the world. :)

 Me too :P

Just try to be around people, they may let you join them :)

:D

My ears? They are always fine :)

Those hard integrals are what I have learnt the days I have temp. left the forum.

I knew how to do them... all of them :D

Hi Max,

Were your ears burning? 

 I just put up a post thanking you and as I wrote it I was wondering where you had got to and all of a sudden you appear !  Just like magic    

Your integrals are  too hard    

Maybe Heureka will answer them. 

Do you already know how to do them or have you looked them up in Wolfram Alpha or an integral calculator?

If I had more energy I'd do that  

It is always nice to see you here :D

\(x=1+\dfrac{1}{2+\frac{1}{1+...}}\\ 1+\dfrac{1}{x}=1+\dfrac{1}{1+\frac{1}{2+...}}\\ \)

First we let x = {1;2,1,2,1,...}, then 1+1/x = {1;1,2,1,2,1,2,...}

Then we find the value of x.

\(x=1+\dfrac{1}{2+\frac{1}{1+...}}\\ x=1+\dfrac{1}{2+x}\\ (x-1)(2+x)=1\\ x^2+x-3=0\\ (x+\dfrac{1}{2})^2=\dfrac{13}{4}\\ x+\dfrac{1}{2}=\pm\dfrac{\sqrt{13}}{2}\\ x=\dfrac{-1+\sqrt{13}}{2}\text{ or }x=\dfrac{-1-\sqrt{13}}{2}(rej.)\\ \)

The value of the original expression is 1+1/x which is:

\(1+\dfrac{2}{\sqrt{13}-1}\\ =\dfrac{\sqrt{13}+1}{\sqrt{13}-1}\\ =\dfrac{14+2\sqrt{13}}{12}\\ =\dfrac{7+\sqrt{13}}{6}\)

:D

~The smartest cookie in the world.

Why can't I see this in the sticky notes?

Given that y is inversely proportional to x² and x is always positive. If y=27 when x = 6, calculate the value of x when y=3

\(y\;\alpha \; \frac{1}{x^2} \qquad x>0\\ y\;= \; \frac{k}{x^2} \\ 27= \; \frac{k}{36} \\ k=27*36\\ k= 972\\ y=\frac{972}{x^2}\\~\\ when \;y=3\\ 3=\frac{972}{x^2}\\ 3x^2=972\\ x^2=324\\ x=18 \)

what term multiplys with 5pq^3, 6p^5q^2 and 4p^2q^4 to give 480. 

\(5pq^3\times 6p^5q^2\times 4p^2q^4 \times \fbox?=480\\ 120p^8q^9 \times \fbox?=480\\ \fbox?=\frac{480}{120p^8q^9}\\ \fbox?=\frac{4}{p^8q^9}\\ \)

Привет

1) The graph above is a function as it passes the vertical line test. 

2) A function is considered to be discontinuous if you cannot outline the entire function with only one stroke of a writing utensil. The discontinuities lie on x=1, x=2, and x=3, assuming the scale of the graph is 1:1. 

At x=1, the type of discontinuity is a jump discontinuity because the linear section of the graph suddenly ends, and a new part of the graph starts again at the same vertical line.

At x=2, that is a removable discontinuity because there is a hole located at (2,0).

At x=3, this is an infinite discontinuity because x=3 is a vertical asymptote. If you identify a vertical asymptote, you have identified an infinite discontinuity.

3) \(\text{Domain:}\hspace{1mm}[0,3)\cup(3,4]\)

\(\text{Range:}\hspace{1mm}(-\infty,+\infty)\)

Find the mean of the integers.

−54,−32,−70,−25,−65,−42

\(\begin{array}{|rcll|} \hline && \frac{-54-32-70-25-65-42}{6} \\ &=& \frac{-(70+25)-(54+42)-(32+65)}{6} \\ &=& \frac{-95-96-97}{6} \\ &=& \frac{-96-1-96-96+1}{6} \\ &=& -\frac{3*96}{6} \\ &=& -\frac{96}{2} \\ &=& -48 \\ \hline \end{array}\)

How many non-empty subsets of \(\{ 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 \}\) consist entirely of prime numbers?

(We form a subset of the group of numbers by choosing some number of them, without regard to order.

So, \(\{1,2,3\}\) is the same as \(\{3,1,2\}\).)

The set of prime numbers \( \{ 2, 3, 5, 7, 11 \}\) consist of 5 elements.

The subsets with one element \( = \binom{5}{1} = 5 \) subsets.
The subsets with two elements \(= \binom{5}{2} = 10\) subsets.
The subsets with three elements \(= \binom{5}{3} = 10\) subsets.
The subsets with four elements \(= \binom{5}{4} = 5\) subsets.
The subsets with five elements \(= \binom{5}{5} = 1\) subset.

The sum is 5+10+10+5+1 = 31 subsets

It's easy! You remove the parentheses and put the appropriate signs. Need a concrete example.

What is 54 to the power of 54

Hello Asad!

\(54^{54}\)

\(=3.542118045010639240328481337533320\\ 71263980863803681247321110974326255\\ 2383710557968252383789056\times10^{93}\)

  !

Calculators are useful for expression like \(\frac{3}{10}\div\pi*45\div3+5*6-3+650*4\approx2628.43239\)

All look fine except for No. 4:

4)

$150 x 0.055 x 4 =$33.00 interest earned over 4 years.

No problem!

1 degree = 60 minutes 

1 minute = 60 seconds 

360 seconds in a degree

I don't know if this answers your question or not :(

Thank you!