You must be able to work out some of them!
There is no point in anyone answering for you if you do not put ANY effort into the question yourself.
Please no one else answer. You will just stop this person from even attempting to learn anything.
Hmm thats incorrect :/ thanks for the effort tho :)
I just muddled my way through this.
Maybe there is a better way to do it.
\((x-(\sqrt2+\sqrt3))\quad \text{is a factor}\\consider\\~\\ (x-(\sqrt2+\sqrt3))(x+(\sqrt2+\sqrt3))\\ =x^2-(\sqrt2+\sqrt3)^2\\ =x^2-(2+3+2\sqrt2\sqrt3)\\ =x^2-(5+2\sqrt6)\\ =((x^2-5)-2\sqrt6)\\~\\ Now\\ ((x^2-5)-2\sqrt6)((x^2-5)+2\sqrt6)\\ =(x^2-5)^2-(2\sqrt6)^2\\ =x^4-10x^2+25-4*6\\ =x^4-10x^2+1\\~\\ so\\ x^4-10x^2+1 \quad has \quad \sqrt2+\sqrt3\quad \text{ as a root} \)
LaTex coding:
(x-(\sqrt2+\sqrt3))\quad \text{is a factor}\\consider\\~\\ (x-(\sqrt2+\sqrt3))(x+(\sqrt2+\sqrt3))\\ =x^2-(\sqrt2+\sqrt3)^2\\ =x^2-(2+3+2\sqrt2\sqrt3)\\ =x^2-(5+2\sqrt6)\\ =((x^2-5)-2\sqrt6)\\~\\ Now\\ ((x^2-5)-2\sqrt6)((x^2-5)+2\sqrt6)\\ =(x^2-5)^2-(2\sqrt6)^2\\ =x^4-10x^2+25-4*6\\ =x^4-10x^2+1\\~\\ so\\ x^4-10x^2+1 \quad has \quad \sqrt2+\sqrt3\quad \text{ as a root}
hi guest!
this problem looks super hard and i have no idea how to solve it, but i think someone else has already solved it here: https://www.quora.com/What-is-the-six-digit-number-N-such-that-when-its-digits-are-reversed-the-result-is-equal-to-4N
How about you start it? You must be able to add something for yourself!
No one else answer over me, thanks.
PS. here's a good youtube video about finding inverses! https://youtu.be/2zeYEx4eTdc?t=505
\(f(x)=\frac{x+2}{x-2}\)
To find the inverse, you can first replace f(x) for y. \(y=\frac{x+2}{x-2}\)
1. we replace the x with y. \(x=\frac{y+2}{y-2}\)
2. cross multiply:
\(x(y-2)=y+2\)
\(xy-2x=y+2\)
\(xy-y=2x+2\)
\(y(x-1)=2x+2\)
3. divide both sides by x-1: \(y=\frac{2x+2}{x-1}\)
And that's the answer! \(\boxed{y=\frac{2x+2}{x-1}}\)
i hope this helped you!
:)
The number of tilings is given by the Fibonacci number \(F_{10} = 55\)
oh ok.. thanks guest! :)
You can prove that statement A, B, D, E, and F are true.
Thats incorrect i think
sorry, I read the problem wrong the first time, the real answer is x≤−√3 or −1 < x ≤ √3
{THIS ANSWER IS WRONG, PLEASE READ MY SECOND ANSWER} You first multiply both sides by (x+1) and you get x+3=x2+x. You then subtract x from both sides and you get 3=x2 which is \(\sqrt{3}\)=x.
The minimum value is 2/3.
The maximum value is 325. There is no minimum value.
This looks hard!
The ranges of quadratic functions are infinite, so there are an infinite number of solutions.
22.5 square units
1234 1357 2345 2468 3456 3579 4567 5678 6789 Total = 9 such numbers.
First number =1000000001
Last number =9999999999
[9999999999 - 1000000001] / 2 + 1 =4,500,000,000 such numbers.
n mod 8 =0 n mod 9 =1 n mod 10 =2
The LCM of 8/2, 9, 10/2 =180
Using "Chinese Remainder Theorem + Modular Multiplicative Inverse" 180 + 172 = 352 minimum number of trees
https://web2.0calc.com/questions/divisibility-remainders
[1000/5 + 1000/5^2 + 1000/5^3 + 1000/5^4] =249 zeros.
(100, 101, 120, 121, 122, 123, 124, 125, 144, 145, 146, 147, 148, 149, 168, 169, 170, 171, 172, 173, 192, 193, 194, 195, 196, 197>>Total =26 such numbers.
