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Factor  [98! + 99!  + 100!] =

2^99×3^46×5^26×7^16×11^8×13^7×17^5×19^5×23^4×29^3×31^3×37^2×41^2×43^2×47^2×53×59×61×67×71×73×79×83×89×97 (240 prime factors, 25 distinct)

As you can see, the largest 5^n =5^26

Some chemistry questions are answered here.....what is your question?

Odell and Kershaw run for 30 minutes on a circular track. Odell(1) runs clockwise at 250 m/min and uses the inner lane with a radius of 50 meters. Kershaw(2) runs counterclockwise at 300 m/min and uses the outer lane with a radius of 60 meters, starting on the same radial line as Odell. How many times after the start do they meet?

Odell und Kershaw rennen 30 Minuten lang auf einer Kreisbahn. Odell(1) fährt mit 250 m / min im Uhrzeigersinn und nutzt die innere Fahrspur mit einem Radius von 50 Metern. Kershaw(2) fährt mit 300 m / min gegen den Uhrzeigersinn und benutzt die äußere Fahrspur mit einem Radius von 60 Metern, beginnend auf der gleichen radialen Linie wie Odell. Wie oft treffen sie sich nach dem Start?

Hello Guest!

\(\alpha= \omega t=\frac{v}{r}\cdot t\\ \alpha_1= \omega_1 t=\frac{v_1}{r_1}\cdot t\\ \alpha_2=2\pi - \omega_2 t=2\pi -\frac{v_2}{r_2}\cdot t\\ \color{blue} \alpha_1=\alpha_2 \)

\(\frac{v_1}{r_1}\cdot t=2\pi -\frac{v_2}{r_2}\cdot t\\ (\frac{v_1}{r_1}+\frac{v_2}{r_2})\cdot t=2\pi\\ \color{blue}t=2\pi/(\frac{v_1}{r_1}+\frac{v_2}{r_2}) \)

\(t=2\pi/(\frac{250m}{min\cdot 50m}+\frac{300m}{min\cdot 60m})=2\pi \times 0,1min\ \times \frac{60sec}{min}=\color{blue}37.7sec\)

\(30\times 60sec /37.7sec=\color{blue}47.7\)

Odell and Kershaw meet 47 times.

  !

Once again, we have an example of someone who doesn't want others to learn from the question. Please do not delete your question after someone has answered it. There might be people out there who want to learn from the question.

\(\begin{array}{rl} &k^2 - 81\\ =&k^2 \color{red} - 9k + 9k\color{black} -81\\ =&(k^2 - 9k) + (9k - 81)\\ =&\color{red}k\color{black}(k-9)\color{red}+9\color{black}(k-9)\\ =&(k - 9)(\color{red}k + 9\color{black})\\ =&(k -9)(k+9) \end{array}\)

(0,2)                                                     (-3, 7)

2(0) + 5(2)  = 10                                  2(-3)  + 5(7) =  10

0  +   10  = 10                                        -6  +  35   =  10                                

10 = 10                                                        29  =  10                                  

Is true  !                                                       Is false !

So  (0,2) lies on the line but not (-3, 7)

If $|x-2|=p$, where $x<2$,  then what is $x-p$ in terms of $p$ ?

Wenn  | x-2 | = p  , wobei  x <2  ist, was ist dann  x-p  in Bezug auf  p ?

Hello Guest!

             x =   | - 3   - 4    - 5   

p = |x - 2| =   |   1     6      7  

        x - p =   | - 4 - 10   - 12  

   ??

3)Determine the range of the function g(x)= 5x^2-2x+1. Enter your answer in interval notation.

This is a parabola that turns upward

The x coordinate of the vertex  is    2/ [ 2 * 5]    =   1/5

The y coordinate of the vertex  is

5(1/5)^2 - 2(1/5) + 1   =

1/5 - 2/5 + 1  =

4/5

So...the range  is   [ 4/5, infinity )

2) Expressing your answer in interval notation, find all values of s such that x(x-2)^2(x+1)<0.

I think you must mean "x", not "s."

x ( x - 2)^2 (x + 1)  < 0

Note that when  x ≥ 0.... the result will be ≥  0

When  -1 < x < 0....the result will be < 0

When -inf < x < -1.....the result will be ≥ 0

So.....this will be < 0   on  the interval   ( -1, 0 )

1) Expressing your answer in interval notation, find all values of k such that the parabola y=x^2 + kx + 11 does not intersect the line y=2.

Let's see where the parabola DOES intersect y = 2

x^2 + kx + 11  = 2

x^2 + kx + 9  = 0

Where the discriminant is ≥ 0, we will have real solutions....so....

k^2 - 4*9  ≥ 0

k^2 ≥ 36

So.....the intervals  where the parabola will intersect the line y = 2  are when k =

(-infinity, -6 ] U [6, infinity )

So.....when    -6 < k < 6      the parabola WILL NOT intersect the line y = 2

Nice, hard work!

By the way, I like your riddles, it helps me warm up my brain and makes me feel mentally healthy lol.

lez draw a 3 by 3 chess board yeet

*amazing chess board*

A knight moves like this (I play chess so its pretty easy for me to visualize)

Since a knight moves one square up and two across, two squares down and one across, two squares up and one across, and other like combinations. (or an L shape)

it cannot move if it was placed in the center square

So there is only eight other squares it can move around.

There are two types, a corner square and a side square

Lets count number of ways you can type 10 digit numbers for a CORNER square.

Lets list number of combinations of ten digit numbers

Digit 1 - 2 ways, because the knight can only move on these directions

Digit 2 - 2 ways

Digit 3 - 1 way, there is only one way to go back.

Digit 4 - 2 ways, starting a pattern

Digit 5 - 1 ways

Digit 6 - 2 ways

Digit 7 - 1 ways

Digit 8 - 2 ways

Digit 9 - 1 ways

Lets calculate. 2 * 2 * 1 * 2 * 1 * 2 * 1 * 2 * 1 = 32

32 * 4 (corner squares) = 128 corner ways.

Now lets start the sides

Digit 1 - 2 ways

Digit 2 - 1 ways

Digit 3 - 2 ways

Digit 4 - 1 ways

Digit 5 - 2 ways

Digit 6 - 1 ways

Digit 7 - 2 ways 

Digit 8  - 1 ways

Digit 9 - 2 ways

so 2 * 2 * 2 * 2 * 2 = 32 ways

32 * 4 = 128 ways

128 + 128 = \(\boxed{256}\) ways

PHEW ! Someone check this, I probably over counted.

11-9 = 2

uses 1 and 9

so 91 

and 19

This is just my weird thought process lol.

I assume this is an AOPS problem, I also assume that you forgot to provide a picture. Whoops!

31  =  21 + 8 + 2                             41  =  34 + 5 + 2

32  =  21 + 8 + 3                             42  =  34 + 8

33  =  cannot be written                  43  =  34 + 8 + 1

34  = 34                                          44  =  34 + 8 + 2

35  = 34 + 1                                    45  =  34 + 8 + 3

36  =  34 + 2                                   46  =  cannot be written

37  = 34 + 3                                    47 =  34 + 13

38  = 34 + 3 + 1                              48  =  34 + 13 + 1

39  = 34 + 5                                    49  =  34 + 13 + 2

40  = 34 + 5 + 1

1)  These are the only numbers that meet all the conditions:

124   126   128   136   139   148   248   Total =  7

Of the final five contestants in a television show, three are female and two are male.  If two are chosen randomly to be the final contestants, what is the probability that both are female? Express your answer as a common fraction.     

Out of the original     five contestants, the odds that a female will be chosen are 3 out of 5  =  3/5

Out of the remaining four contestants, the odds that a female will be chosen are 2 out of 4  =  2/4

The odds of both the above happening are 3/5 times 2/4  =  6/20  =  3/10

.

295/396

The total  possible  outcomes  =  C(5,2)  = 10

The number of outcomes that have two females  is  C(3,2) * C (2, 0)  = 3 * 1  = 3

So....the probability  =

3  / 10

We have 2^3 * 2^2  =  32 outcomes

The  ones where Coin A has more heads outcomes than Coin B  are

  A         B

HHH   HH      =  1* 1  =  1 way

HHH   TH    =    1 * 2   = 2 ways

HHH   TT    =      1  * 1   = 1 way

HHT    HT  =       3 * 2  = 6 ways

HHT   TT   =       3 * 1  =  3 ways

HTT   TT   =       3 * 1  =  3 ways

So....the total ways  =  16

So....the probability  =  16 / 32    =  1  / 2

2 )  Here's what I get  for "n"

                        n

3 * 5   * 1   =   15

3 * 5  * 4  =     60

3 * 5 * 9   =   135

3 * 5 * 16  = 240

3 * 5 *  25  = 375

3 * 5  * 36  = 540

3 * 5 * 49 =  735

3 * 5 * 64 =  960

-3   added to numerator   2

-3  added to denominator 5

2/5 = .4

(5+x) / (8+x) = 4/10

5+x = (8+x) (4/10)

5 + x = 32/10 + 4/10x

1.8 = -6/10 x

1.8 (-10/6) = x

-18/6 = x

x = -3