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Compute 997^ -1  modulo 1000. Express your answer as an integer from 0 to 999.

997^(-1) mod 1000 ==333

Thanks, much appreciated.

Hi Melody.

Here's the difference table that you asked to see.

\(\displaystyle \begin{array}{ccccccc}x&f(x)&\Delta &\Delta^{2} &\Delta^{3} &\Delta^{4} &\Delta^{5} \\ 1 & 2 &&&&&&\\ &&1 &&&&\\ 2&3 &&0&&&\\ &&1&&0&&\\ 3&4&&0&&0&\\ &&1&&0&& -140 \\ 4&5 &&0&& -140 &\\ && 1 && -140 &&\\ 5&6 && -140 &&&\\ && -139 &&&&\\ 6& -133 &&&&&\\ \end{array}\)

The x = 1 and f(x) = 2 at the top of the first two columns are x0 and f(x0).

Running down diagonally from the 2 are the forward differences for f(x0). They are, 1, 0, 0, 0, -140.

The next diagonal 1, 0, 0, -140 running down from the 3 are the forward differences for f(x1) and so on.

Let me know if you need any more info.

Tiggsy.

Hi guest,

My original answer (the one you referred to) does look at the multiplication of 2  two digit numbers 

The header row is the units digit of the first number 

The header column is the units digit of the second number.

What I did not take into consideration is that the numbers are supposed to be different from one another.

This makes the question much more difficult.  Each cell will no longer occur an equal number of times.

So I have redone it.  I think this method is right but I am less confident than i was with the original answer.

A typo: 4 / 100 = 1 / 25

Hi Melody: I think they want "Two different 2-digit numbers are randomly chosen and multiplied together."

So the probability will be  4/100 = 0.04     (I have fixed a typo)  Thanks guest  

This is a super easy proof.

Hint:    \(Cos(2A)=cos^2A-sin^2A\)

As follows:

x*(5/7)=20
20/(5/7)=x

What is the radius of the circle inscribed in triangle ABC if AB=22, AC=12, and BC=13? Express your answer in the simplest radical form.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

1/   Find semiperimeter       s = (a + b +c) / 2

2/   Find the area of ΔABC       A = sqrt[ s(s - a)(s - b)(s - c)]    

3/   Calculate the radius     r = A / s 

There are: 90 nCr 2 =4005 possible products.

Only 153 of them end in "1"

Therefore, the probability = 153 / 4005 = 17 / 445

Please check your work, this is incorrect

Since f(x) is divisible by x^3, f(x) is of the form ax^5 + bx^4 + cx^3.

You then want ax^5 + bx^4 + cx^3 + 2 to be divisible by (x + 1)^3.  Using long division, you get the equations

-10a  + 6b - 3c = 0

4a - 3b + 2c = 0

-a + b - c + 2 = 0

==> a = 6, b = 16, c = 12

So f(x) = 6x^5 + 16x^4 + 12x^3.

I could find only 2 values for x which give an integer solution for every value of n:

That is when x =1   and    x = 3

There are: 100 nCr 2 =4,950 possible sums

There are 4243 sums that are NOT divisible by 7, which is also the size of the largest subset

Find the equation of the line that passes through the point (5,8) and has a slope of 2 in slope point form.

Hello Guest!

\(y=2(x-5)+8\)

\(y=2x\)-2

  !

a==[3,  27,  51,  75,  99 ]>Total = 5 such integers

(y-8) = 2 (x-5)               that's it !

Via Pythagorean Theorem

  (a + 1)² + (2a + 1)^{2  =} 36 a²

  5a² + 6a + 2   =  36 a²       re-arrange

-31a² +6a +2 = 0                 Plug into Quadratic Formula     a = -31    b = 6     c = 2

      to find    a = .368585

\(\text{PLZ help!}\)

Perimeter    P = √144 + √32 = 17.65685425

A, B, and D statements are true.

Angle DAE = 150 degrees