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Here are the last 10 non-zero digits of 2019! =2,587,693,056. So, the last two digits multiplied =5 x 6 = 30.

If you meant added together, then: 5 + 6 = 11. Take your pick !!.

Asked again here:

I answered this question recently, or something very similar.  Try searching my answer.

Lance has a regular heptagon (7-sided figure). How many distinct ways can he label the vertices of the heptagon with the letters in OCTAGON if the N cannot be next to an O ? Rotations of the same labeling are considered equivalent.

Here, I search for you, it was easy to find.

It might be worth noting BC that I do not answer questions where the asker has not gone to any effort to present it properly.

This may not matter because a number of other people like forensic mathematics.  I am just letting you know.

I have no idea either CoolStuffYT    

a=9 and b=4

1. We can use the sine law to find P1 P2

By the sine law, P1 P2 = sin 15.  We can also find P1 P_3 = sin 60, P1 P4 = sin 90, etc.

So (P1 P2)^2 + (P1 P3)^2 + … + (P11 P12)^2 = 12 (sin 15)^2 + 12 (sin 30)^2 + 12 (sin 45)^2 + … + 12 (sin 90)^2 = 42.

2.  Arc WY - Arc XV = Angle U = 30 degrees.

Arc WY + Arx XV = 360 - 170 - 110 = 80 degrees.

Therefore, arc WY = (30 + 80)/2 = 55 degrees.

V + F - E = 2       Leonhard Euler

V + 30 - 62 = 2

        V - 32 = 2

               V = 34

Three choices for first integer    6   7    8

   second integer has 4 choices of those remaingn on the list

     third integer has 3 choices

      last integer 2 cohoices

3 x 4 x 3 x 2  = 72 integers >6000

The possible values are -1, 1, and 2.

There are 7! = 5040 ways to label the heptagon.  There are 4! = 24 ways where the O's are next to an N, so the number of ways where N is not next to an O is 5040 - 24 = 5016.

You can make: 5! =120 integers > 6000 such as: {3, 5, 6, 7, 8} | {3, 5, 6, 8, 7} | {3, 5, 7, 6, 8} | {3, 5, 7, 8, 6} | {3, 5, 8, 6, 7} | {3, 5, 8, 7, 6} | {3, 6, 5, 7, 8} | {3, 6, 5, 8, 7} | {3, 6, 7, 5, 8} | {3, 6, 7, 8, 5} | {3, 6, 8, 5, 7} | {3, 6, 8, 7, 5} | {3, 7, 5, 6, 8} | {3, 7, 5, 8, 6} | {3, 7, 6, 5, 8} | {3, 7, 6, 8, 5} | {3, 7, 8, 5, 6} | {3, 7, 8, 6, 5} | {3, 8, 5, 6, 7} | {3, 8, 5, 7, 6} | ... (total: 120)

P.S. You have 5 numbers, so ANY combination will be > 6000 !!. Unless you meant > 60,000 then: 120 / 5 = 24 integers that start with each digit. So, if you remove 3 and 5 from the beginning of each integer, then you will have:  120 - [2 x 24] = 72 integers > 60,000.

There are 7!/7 ways to label a heptagon, not 7! . You are overcounting.

Area shaded region of the triangle ABC is 80 sq. units, what is the area of triangle ABC?

Solve sqrt(4x - 3) + 10/sqrt(4x - 3) = 7.

See https://web2.0calc.com/questions/help_99449.

let the last vertex of the shaded triangle be y

Triangle BYC area= 1/2bh (General formula) 

1/2(2n)*h=80

2n*h=160

2nh=160

nh=80

Consider triangle ABC

area=1/2(3n)*h

1/2*3nh

we know nh=80

1/2*3(80)=120

so triangle ABC area=120

Also triangle ABY=120-80=40 

Thanks Melody. I wish I was as good as you are at these combinations and permutations. Bravo!

Music is great if you have a passion for it.   I expect your parents are very passionate about it.

Computer science is more practical for most peope.

I was 'made' to do music for many years when I was young. It was not the best use of my time but even so I have never been sorry that i learned it. 

Music is like mathematics in that it is a language of its own. I do not play an instrument any more but I am still glad that I have some understanding of how music is constructed.  I am glad I can still more or less read music and understand some musical terms. This helps me to appreciate the actual muisic more.

Talk to your parents, maybe you can reach a compromise and do music in some less intense, more recreational, way.

Thanks guest,

It is always very helpful when someone gives the answer via computer code.

Even better still when the method used has been stated clearly.   

It is not that hard to do it mathematically.

I just looked at how many triplets would add to 10.  There is only 6 of them

\( 6,3,1,1,1 \qquad \qquad \frac{5!}{3!}=\frac{120}{6}=20 \;ways \\ 6,2,2,1,1 \qquad \qquad \frac{5!}{2!2!}=\frac{120}{4}=30 \;ways \\ 5,4,1,1,1 \qquad \qquad \frac{5!}{3!}=\frac{120}{6}=20 \;ways \\ 5,3,2,1,1, \qquad \qquad \frac{5!}{2!}=\frac{120}{2}=60 \;ways \\ 4,4,2, 1,1 \qquad \qquad \frac{5!}{2!2!}=\frac{120}{4}=30 \;ways \\ 4,3,3 ,1,1 \qquad \qquad \frac{5!}{2!2!}=\frac{120}{4}=30 \;ways \)

20+30+20+60+30+30 = 190

Altogether there are 6^5 = 7776 combionations

190/7776 = 0.0244341563786008

Just as guest's computer program already computed.

Circle's radius      >  r = 2

Rhombus side      >  s = r

--II-- long diagonal > d₁ =?             d₁ = sqrt[s²- (r/2)²]*2      d₁ = 3.464

--II-- short diagonal > d₂ = r           

Rhombus  area       > A = ?             A = 1/2(d₁*r)  or  A = s * sqrt(3)  or A = (d₁*d₂)/2   

                                                                                      A = 3.464 u²  

Triangle side  >  s = 2 

---II---   angle  >  q = 60°  

---II---  height  >  h = ?                   h = tan(60°)  = 1.732      

Square side    >  a = 2

Total height     >  H = ?                  H = a+h      H = 2+1.732     H = 3.732              

Desired angle >  A = ?                   tan(q/2) = (a/2)/H  = 15        A = 15*2   = 30°