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The engine has to be first, then the eight cars can be arranged in eight factorial ways...

...unless the caboose has to be last, and then the answer is seven factorial. 

8! is 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40,320 

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thx Guest

how do i become a moderator 

Get several thousand hearts, then come back and ask that question. 

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ok thx Question

You do know...and tbh I can't say I'm an expert, but one Cannot just stroll in Become the alfa wolf. Moderator is a buissness for only the smartest of us...with myself excluded of course.

But I'm no expert. Just ask any of the already Moderators and I'm sure they'll let you know.

a quadralateral has no pairs of parallel sides  which two shapes can it be

i got a kite and trapezium 

but doesnt look right

Trapezium is correct if you're in the right country. 

I copied this definition from the internet:  Trapezium, describing a geometric shape, has two contradictory meanings: Outside the US and Canada: a quadrilateral with at least one pair of parallel sides (known in the US as a trapezoid) In the US and Canada: a quadrilateral with no parallel sides (known elsewhere as a general irregular quadrilateral) 

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Formula  for arc length  = 

arc length  =  r  θ         where θ is in radians

So  we  have

2pi  =  5 θ             solve for θ  and we have that

θ  =  [ 2 pi ]  / 5

The area  of the sector   =

(1/2) r^2 θ   =   (1/2)(5^2) (2pi) / 5   =   5 pi  in^2  ≈ 15.7  in ^2

The rectangular  shape  has more area  = (5 * 4)   = 20 in ^2

Just use the quadratic formula.

The bit under the square root is a perfect square.

If you need help you can ask but you must show a proper contribution of your own first.

Please no one provide an exact answer. We should strive for this to be a teaching/help site. Not a cheat site.

guys helppp this is hurting my brain

x - y  = sqrt (3)

x + y =  2

Square both sides  of these

x^2  - 2xy  + y^2   =   3

x^2  +2xy   + y^2  =   4           subtract the first equation from the second and we have that

4xy  =  1            divide both sides by  4

xy  = 1/4

Can't figure this without knowing the arc length......

OK.....sorry, Melody!!!!

Second one

Need to be  careful reading the number  of pupils.....each vertical "tick"  mark represents (1/2)  of a pupil  [ NOT  1  pupil ].......this seems to be  an odd way of presenting this...!!!!!

a)   number scoring  more than 8  =   1 + 4  =  5

b)   number of students  =   3 + 5 + 2  + 6 + 12 + 1 + 4   =  33

c)  We are assuming  that the whole class took the test  (no absentees)

How about you explain?  If you have tried this exercise then you must have some idea. 

What is a tesselation?

Watch this:

First one

a)   the triangle  has two equal angles....therefore...it is isosceles

b)   acute.....it is  < 90°

pls help xD

999  =  

111 *  9  =

(37 * 3) * (3^2)  =

37 * ( 3)^3

The prime factors are 37  and 3

Their sum  is   40

tan (x)   =  - 2

We are  in the second quadramt.....so this implies  that   (x,y)   = ( -1, 2)

cos 2x  =   1 - 2sin^2 x

sin (x )  =  2  / sqrt [(-1)^2  + 2^2  ]  =   2/sqrt (5)

So....sin^2  (x)  =    4/5

Then  cos 2x  =     1  - 2(4/5)  =   1  -8/5  =     -3/5

The eight train cars can be arranged in  8 !  ways  = 40,320  arrangements

Sorry....I don't know anything about this topic......maybe someone else on here can help....

plz help it says i got them wrong dont know how plz explain

Sorry for the late reply.

We choose the two paired ranks in \(\binom{13}{2} = 78\) ways and the remaining rank in \(\binom{11}{1} = 11\) ways.

Now you know this calculate the number of ways you can pick the suit.

Good luck!

Your effort to get that 1 point is commendable though.

Me: That's just 1 point. Meh.

1/7 = 0.142857142857...