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Perimeter  P = 2*10pi

What is the remainder of \(N=1\times 3\times5\times7\times...\times101\)  when it is divided by 8?

So
\(\begin{array}{rcll} N &\equiv& x \pmod{8} \\ 1\times 3\times5\times7\times\ldots\times101 &\equiv& x \pmod{8} \\ \end{array}\)

\(\begin{array}{|rcl|rcl|rcl|rcl|} \hline \mathbf{1} &\equiv& 1 \pmod{8} & \mathbf{3} &\equiv& 3 \pmod{8}& \mathbf{5} &\equiv& 5 \pmod{8}& \mathbf{7} &\equiv& 7 \pmod{8} \\ \mathbf{9} &\equiv& 1 \pmod{8} & \mathbf{11} &\equiv& 3 \pmod{8}& \mathbf{13} &\equiv& 5 \pmod{8}& \mathbf{15} &\equiv& 7 \pmod{8} \\ \mathbf{17} &\equiv& 1 \pmod{8} & \mathbf{19} &\equiv& 3 \pmod{8}& \mathbf{21} &\equiv& 5 \pmod{8}& \mathbf{23} &\equiv& 7 \pmod{8} \\ \mathbf{25} &\equiv& 1 \pmod{8} & \mathbf{27} &\equiv& 3 \pmod{8}& \mathbf{29} &\equiv& 5 \pmod{8}& \mathbf{31} &\equiv& 7 \pmod{8} \\ \vdots &&&\vdots&&&\vdots&&&\vdots \\ \mathbf{89} &\equiv& 1 \pmod{8} & \mathbf{91} &\equiv& 3 \pmod{8}& \mathbf{93} &\equiv& 5 \pmod{8}& \mathbf{95} &\equiv& 7 \pmod{8} \\ \mathbf{87} &\equiv& 1 \pmod{8} & \mathbf{99} &\equiv& 3 \pmod{8}& \mathbf{101} &\equiv& 5 \pmod{8} \\ \hline \end{array} \)

\(\small{ \begin{array}{|rcll|} \hline (1\times 3\times5\times7)\times (9\times11\times13\times 15)\times (17\times19\times21\times 23)\times \ldots \times (89\times91\times93\times 95)\times(97 \times99 \times101) &\equiv& x \pmod{8} \\ (1\times 3\times5\times7)\times (1\times 3\times5\times7)\times (1\times 3\times5\times7)\times \ldots \times (1\times 3\times5\times7)\times(1 \times3 \times5) &\equiv& x \pmod{8} \\ \boxed{ (1\times 3\times5\times7) = 105 \equiv 1\pmod{8}\\ (1\times 3\times5) = 15 \equiv 7\pmod{8} } \\ (1)\times (1)\times (1)\times \ldots \times (1)\times(7) &\equiv& {\color{red}7} \pmod{8} \\ \hline \end{array} }\)

The remainder of \(N=1\times 3\times5\times7\times\cdots\times101\)  when it is divided by 8 is \(\mathbf{\color{red}7}\)

In triangle ABC, AB = BC = 25 and AC = 40. What is sin∠ACB?

Let M be a midpoint of AC

BM = sqrt( BC² - MC² )

sin∠ACB = BM / BC

There are 3^n - 2^n possible sequences.

In triangle ABC, AB = BC = 25 and AC = 40. What is \(\sin( \angle ACB)\) ?

Heron's formula states that the area of a triangle whose sides have lengths \(a \), \(b\), and \(c\) is

\(\qquad A = \sqrt{s(s-a)(s-b)(s-c)}\),

where s is the semi-perimeter of the triangle; that is,

\(\qquad s=\dfrac{a+b+c}{2}\)

\(\begin{array}{rcll} \text{Let $BC=a=25$} \\ \text{Let $AC=b=40$} \\ \text{Let $AB=c=25$} \\ \end{array}\qquad \begin{array}{|rcll|} \hline \mathbf{ s } &=& \mathbf{ \dfrac{a+b+c}{2} } \\ s &=& \dfrac{25+40+25}{2} \\ \mathbf{ s } &=& \mathbf{45 } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{ A } &=& \mathbf{\sqrt{s(s-a)(s-b)(s-c)} } \\ A &=& \sqrt{45(45-25)(45-40)(45-25)} \\ A &=& \sqrt{45*20*5*20} \\ A &=& \sqrt{15^2*20^2} \\ \mathbf{ A } &=& \mathbf{15 * 20} \\ \hline \end{array}\)

Formula: \(2A = ab\sin(\angle ACB)\)

\(\begin{array}{|rcll|} \hline 2A &=& ab\sin(\angle ACB) \quad | \quad \mathbf{ A = 15 * 20},\ a=25,\ b=40 \\ 2*15 * 20 &=& 25*40 \sin(\angle ACB) \\ 15 * 40 &=& 25*40 \sin(\angle ACB) \quad | \quad : 40 \\ 15 &=& 25 \sin(\angle ACB) \\ \sin(\angle ACB) &=& \dfrac{15}{25} \\ \mathbf{ \sin(\angle ACB) } &=& \mathbf{\dfrac{3}{5}} \\ \mathbf{ \sin(\angle ACB) } &=& \mathbf{0.6} \\ \hline \end{array} \)

x² + 3x + 20 = 360

If x = 17 then    m / n = 11 / 79  

LON  + MON   form a straight line = 180 ^o

8x-13     +      7X-17   = 180       Solve for 'x'      then    MON  =   8 x - 13

For 8 = 1 permutation

For 16 =6,147 permutations

For 24 =98,813 permutations

For 32 =98,813 permutations

For 40 =6,147 permutations

For 48 =1 permutation

Sum them all up [1 + 6,147 + 98,813 +98,813 +6,147 + 1] =209,922 / 6^8 =34,987 / 279,936

The answer is \(4 \times (\frac{25 \pi}{2} - 25) = \boxed{50 \pi - 100}\)

So, each petal shaped figure is area of 2 quarter circles of length 5 - the area of a square with side 5

If DG is at most 1, we can draw a circle with radius 1 and center D.

We know the circle and triangle will share a 60 degree sector.

So the area of the sector is pi/6.

The area of the entire equilateral triangle is sqrt3/4*a^2, and since a=3, 9sqrt3/4.

3.8971 is the approximate area of the triangle, and 0.5236 is the approximate area of the sector.

So 0.5236/3.8971 which is around 0.134356

-tigernathan

9 tile diagonal total is 5 tiles each direction    5 x 5 = 25 tiles

thank you! yeah I forgot about adding a-b=8 and a+b=8 together. thank you so much for the work you put into the clear solution! :)

thank you so much!!! i see what i did wrong now. thank you for the great and clear solution!!! :)

I agree.

 If the last three digits of a whole number are divisible by 8, then the entire number is divisible by 8. (By googling)

I don't have an effective way to determine a 3 digit number wehether is multiple of 8 or not.

So I list all 3 digit number consist only numbers 1-6,which are

112,136,144,152,216,232,224,256,264,336,344,312,352,416,456,424,464,432,536,544,512,552,616,624,632,664,656

27 numbers divisible 8 in total  (I did that by adding multiple of 40, so it is  a bit out of order.)

I also wrote a stupid Java program to check my answer.

public static void main(String []args){
        int c=0;
        for (int i = 1;i<=6;i++){
            for (int j = 1;j<=6;j++)
            {
                for(int k = 1;k<=6;k++)
                {
                    int n=i*100+j*10+k;
                    if(n%8==0)
                    {
                        c++;
                    }
                }
            }
        }
        System.out.println(c);
     }

The first five dight can be any integer in {1,2,3,4,5,6}.Therefore, the probability that the number formed is a multiple of eight is \(\frac{6^5*27}{6^8}=\frac{3^3}{6^3}=\frac{1}{2^3}=\frac{1}{8}\).

PS: I believe there is a way to do this problem in abstract algebra, which I rusted at.

https://web2.0calc.com/questions/how-to-solve_20#r1

the point is probably to help you solve it, so if you can dof it another way, that would work as well as long as you get the right answer

but this is the addition method:

1st: Line up the variables in both equation, x above x, y above y, = above =, constant above constant. 2nd: (Optional) Multiply one or both equations so the coefficients of one variable are opposites. 3rd: Add the equations to eliminate one variable. 4th: Solve.

I would prove this with induction.  Have you done induction yet?

Inspirational, how do we realize this?

Jimkey, how do we know this?

a)

\(\text{Prove that   }  n^3 = n + 3n(n - 1) + 6 \binom{n}{3}\\ \text{by counting the number of ordered triples (a, b) of positive integers}, \\\text {where }   1 \le a,b,c \le   \text{in two different ways.}\)

So far I do not understand what this question is on about.

You are asked to prove something, where the only letter used, is n

then a, b and c are introduced to use in the solution.  Where did a, b and c come from?

\text{Prove that   }  n^3 = n + 3n(n - 1) + 6 \binom{n}{3}\\
\text{by counting the number of ordered triples (a, b) of positive integers}, \\\text {where }   1 \le a,b,c \le   \text{in two different ways.}

The conjugate of  a + b  is  a - b  and the conjugate of  a - b  is  a + b.

To simplify this problem, multiply both the numerator and the denominator of the fraction by

the conjugate of the denominator:  sqrt(5) + sqrt(2).

[ sqrt(5) + sqrt(2) ] / [ sqrt(5) - sqrt(2) ]  · [ sqrt(5) + sqrt(2) ] / [ sqrt(5) + sqrt(2) ]

=

[ sqrt(5) + sqrt(2) ] · [ sqrt(5) + sqrt(2) ]  /  [ sqrt(5) - sqrt(2) · [ sqrt(5) + sqrt(2) ]

=

[ 5 + 2·sqrt(5)·sqrt(2) + 2 ] / [ 5 - 2 ]

=

[ 7 + 2·sqrt(10) ] / 3

I have started by finding these three cases

1. a, b and c are equal. Let's name this case, case 1
2. two out of the three integers are equal. (example could be a and b are equal, but c is different, etc.) Let's name this case, case 2
3. a, b, and c are different. Let's name this case, case 3

2d = 17e -8

2e = d-9     d-9 = 2e     multiply by -2      -2d+18= -4e     and add to first equation to get

18 =13e-8

13e = 26

e = 2

me too! just a little hard to follow for my small brain. :) :P