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Cphill answer below..... I neglected the 'distinct' portion of the question

  so it should be :

MATH     can sit in 4 positions

then there are 3 positions left , one has to be a number     10 choices

the other two positions can be a letter    22 x 21     (since   M   A   T   H  are already used)

   but the number can be in one of three positions     x 3

4  x 10  x 22 x 21  x 3      = 55440

Guest 21 mins ago

s/4    -2t    = 2    multiply through  by   4 ⇒   s  - 8t  = 8 ⇒  s  = 8 + 8t

Sub this into the second equation for s

4t  - 5(8+ 8t)    =   9   simplify

4t  - 40  - 40t  =  9

-36t =  9 + 40

-36t =  49

t = -49/36

And s  = 8 + 8(-49/36)    =  -26/9

Multiply first equation by 2  to get

s/2 - 4t = 4     add to second equation to get

-4 .5 s = 13

s = - 2.89     ( or  -  26 /9)

Use this value of s in one of the equations to calc 't'

(-26/9)  / 4  - 2t = 2

t = -98/72   = - 49/36

Thank you CPhill! It really helps out a lot!

Note....for any plate..... the word "MATH" can appear in  any one of 4 positions

Since  the  characters are distinct, I'm assuming that  no letter can be repeated

In the other 3 positions, we  can choose  any of  22 remaining letters * any of 21 remaining letters * 10 digits  and these can be arranged in 3!  = 6 ways

So.....in this scenario, we have

4 * 22 * 21 * 10 * 3    =

55440 possible plates

Ok, so I checked the answer and it was 55,440, although I still don't know how to get it. 

MATH     can sit in 4 positions

then there are 3 positions left , one has to be a number     10 choices

the other two positions can be a letter    26 x 26

    but the number can be in one of three positions     x 3

4  x 10  x 26 x 26 x 3      = 81120   (?)

See this:

https://web2.0calc.com/questions/help_98900

First there is 36 choices because 6*6=36

So now you can count the ways to get 8.

2 and 6

6 and 2

3 and 5

5 and 3

4 and 4

So you get 5/36.

Base 2 palindrome:   10001       = 17 in base 10

Base 16 palindrome:   11           = 17 in base 10        = 10000 in base 2

Actually the answer is 8.

Perimeter of ΔABC = 18 units and the area = 12 square units.

AB = 8

AP = 8.4641

BP = 1.5359

AB + AP + BP = 8 + 8.4641 + 1.5359 = 18

Area of ΔABP = 6 u²

CP ≈ 4.582456 units

I started with 

\(f(x)=ax^3+bx^2+cx+d\)

d is super easy to find

then I formed simultaneous equations  which will be used to find a,b and c

  You do this by substitution

and then solved them using matrices. 

Do what you can and get back to us - show us what you have done.

I already answered this.

If you are not happy with the answer say so on the original thread.

Please give the address of the original post.

That is a neat answer Jugoslav.   

Thanks, I was having a mental block.   ​

The only thing you forgot is that it has to be shown that AB has a perpendicular distance to (0,0) of 3 units.

This is not hard to do, you can just use the Pythagorean triad 3-4-5  but it does have to be shown before your part of the calculations.

Excellent, Guest   !!!!

****

Call the part of  room  that  Ryan can paint in the first 20 minutes,  1/x

When Leonardo  joins in   the part that is painted between 9:20 and 9:40 (20 minutes) is   (1/x) + (2/x)   =  3/x

So  in 40 minutes  the whole job is done....so

(1/x + 3/x)  =  4/ (x)   = 1    ⇒   x  = 4

So....the part painted  between 9  and 9:30   is

(1/4)  +  ( 1/2) (3/4)   =

(1/4)  + 3/8  =

2/8  + 3/8  = 

5/8

Let A  =  (0,6)   B  = (0,0)   and C = (6,0)

AC   =    6sqrt (2)

Triangle   ADE  is similar to  triangle ABC

Call the  side of the square, S =  DE

So

AD/ DE  = AB/BC

AD /S  = 6/6

AD  = S

And triangle   CFG is similar to triangle CBA

So

FG/ CF  = BA/ CB

S / CF  = 6/6

CF = S

So

AD + DF  + FC   =  6sqrt (2)

S  + S + S    = 6sqrt (2)

3S  = 6sqrt (2)

S  = 2sqrt (2)  =  sqrt(8)  cm

Area of square  = 8  cm^2

Square side is a

2(a / tan45º) + a = √72           a = √8           A = 8 cm²

Dividing this into 9 segments, we find that the square takes up 4 segments. 4/9 * the area of the triangle = 4/9*18=
 

8

Using the Law of Cosines

80 - 13 - 61   = -2(sqrt (793)  cos x

6 / [ -2 sqrt (793) ]   = cos x =   -3 /sqrt (793)

sin x   = sqrt  [ 1 - (-3/ sqrt (793))^2  ]   =  sqrt [ 1  - 9/793 ] =  sqrt [ 784/793  ]  = 28/ sqrt (793)

Area  =  (1/2) ( 13) (61) sin x   = 

(1/2) (793) (28/ sqrt (793) )   =

14sqrt (793)  units^2