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sumfor(n, 0, 100, 9^n) =2.988157375 E+95 - This number ends in 501

To finish:

First, eliminate C from the last two equations:

-6A + 2B + C  =  0     --->   x 5   --->   -30A + 10B + 5C  =  0

9A - 15B - 5C  =  1   --->                        9A - 15B  - 5C  =  1

(add down)                                          -11A  - 5B            =  1

Combining this result with the first equation:

     A +  B  =  0   --->   x 11   --->   11A + 11B  =  0

-11A - 5B  =  1   --->                    -11A  -   5B  =  1

(add down)                                               6B  =  1   --->   B = 1/6

A + B  =  0   ---->   A + (1/6)  =  0   --->   A  =  -1/6

-6A + 2B + C  =  0   --->   -6(-1/6) + 2(1/6) + C  =  0   --->   1 + 1/3 + C  =  0   --->   C  =  -4/3

sumfor(n, 1, 123, n^5) =591,309,561,876

First of all, note that all integers are either 0,1, or 2 modulo 3 (if you're not familiar with this terminology, it means that every integer is either a multiple of 3, or it is 1 or 2 away from a multiple of 3).

So, we can think of our numbers as

x | x mod 3

0 0

1 1

2 2

3 0

4 1

5 2

6 0

7 1

8 2

In order to make sure that the sum of any three adjacent numbers is divisible by 3, we have to make sure that any group of 3 three adjacent numbers contains a 0, a 1 and a 2. This is possible only if we arrange our 9 numbers in 3 groups of 3 numbers containing 0,1 and 2 exactly once, repeating always the same pattern.

For example, we could arrange our numbers following the pattern

0,1,2,0,1,2,0,1,2

or

2,0,1,2,0,1,2,0,1

We have 3! = 6 possible patterns. Suppose for example that we choose the pattern

0,1,2,0,1,2,0,1,2

One possible way of following this pattern would be the arrangement

3,1,2,6,4,5,9,7,8

In fact, we substituted every '0' with a multiple of 3 (3, 6 or 9), every '1' with a number 1 away from a multiple of 3 (1, 4 or 7) and every '2' with a number 2 away from a multiple of 3 (2, 5 or 8).

This means that, once we fix a patter, we have 3 choices for the first 3 slots, 2 choices for the next 3 slots, and the final slot will be fixed. So, we have

3*3*3*2*2*2 = 216

possible ways of following a fixed pattern. Since the number of patterns was 6, we have

216*6 = 1296 possible arrangements.

GCF(5!, 10!, 15!) = 5! =120

Angle PBC = 24 degrees.

thx for your help!

The sum of the first n integers is given by n*(n+1)/2 

So you have   n*(n+1)/2 = 10*n

Solve for n.

Correct.

so is it 0?

The sum of the coefficients of a polynomial is just given by the sum of all the terms when x = 1.  So to find it just set x = 1 in your expression.

Like so:

Two lines pass through the point (–6, 7) and each is a distance of 2 from the origin at their closest to the origin. What is the sum of the slopes of these lines? 

Find the circumference of a circle with diameter, dd = 1.17m.
Give your answer rounded to 2 DP.

\(C=d\times \pi =1,17m\times 3,1415926\approx 3.68\)m

  !

Rewrite l_1 in the form  y = -(3/4)x - 14/4  Its slope is -3/4, which means the slope of a line perpendicular to it has slope m = +4/3.

Thus, line L_2 can be written as  y = 4x/3 + b

If it goes through point (-5, 7) then we must have  7 = 4*(-5)/3 + b.

Can you take it from here?

The question implies the result is true for all such x and y, so we could choose x = 2, which has remainder 2 when divided by 5, and y = 4, which has remainder 4 when divided by 5.  Hence x + y = 6, which has remainder 1 when divided by 5, so z = 1.  Therefore (2z - 5)/3 = -1.

There's a difference between working with two decimal places and displaying with two decimal places.  It's dangerous to do the former!  Errors build up quickly.  To display x to two decimal places you can do:  round(100x)/100.

Compute (1^2 + 2^2 + 3^2 + ... + 10^2)/770.

In general

\(\Sigma_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}\)

When n = 10 we have: 

\(\Sigma_{k=1}^{10}k^2=\frac{10.11.21}{6}=385\)

And 385/770 = 1/2

Angle  x = 70 degrees

∑[ (n!) , n, 1, 200] mod 14 = 5

Each side of an equilateral triangle of perimeter 54 is divided into six equal parts.

If the area of the blue triangle is sqrt(3)*n, then find the value of n.

n = A / sqrt(3) = 24  

The surface area of what????

In a 100 meter race, Allen and Bruce reached the finish line in 12 seconds and 15 seconds, respectively.
When Allen finished the race, how far was Bruce from the finish line?
Assume that they each move at a constant rate.

\(v_{\text{Bruce}} = \dfrac{100\ m}{15\ s} \)

\(\begin{array}{|rcll|} \hline s_{\text{Bruce}} &=& \dfrac{100\ m}{15\ s} \times (15\ s -12\ s) \\\\ s_{\text{Bruce}} &=& \dfrac{100\ m}{15\ s} \times (3\ s) \\\\ \mathbf{s_{\text{Bruce}}} &=& \mathbf{20\ \text{meter}} \\ \hline \end{array}\)

Bruce was \(\mathbf{20\ \text{meter}}\) from the finish line.

When a natural number x is divided by 5, the remainder is 2.
When a natural number y is divided by 5, the remainder is 4.
The remainder is z when x+y is divided by 5.
Find the value of \(\dfrac{2z - 5}{3}\).

\(\begin{array}{|lrcll|} \hline & x & \equiv & 2\pmod{5} \qquad (1) \\ & y & \equiv & 4\pmod{5} \qquad (2) \\ \hline (1)+(2): & x+y & \equiv & 2+4\pmod{5} \\ & x+y & \equiv & 6 \pmod{5} \\ & x+y & \equiv & 6-5 \pmod{5} \\ & x+y & \equiv & {\color{red}1} \pmod{5} \\ \hline & z &=& 1 \\\\ & && \mathbf{\dfrac{2z - 5}{3}} \\\\ & &=& \dfrac{2*1 - 5}{3} \\\\ & &=& \dfrac{-3}{3} \\\\ & &=& \mathbf{-1} \\ \hline \end{array}\)