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The y-intercept is \((-10,0)\).

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I got it, Thank you 

10!  and  any greater integer factorial   will always end in two zeroes

So.....

5! =        120

6! =        720

7!  =     5040

8!  =   40320

9! =  362880

So......  the sum has the last two digits =     20 + 20 + 40 + 20 + 80  =   _80

Ohhhh I got it!!!

Thank you CPhill and Melody I think I understand it now. 

The magnitude of a point on the complex plane is just the distance of that point to the 0.   (the origin)

That is why the symbol used is the same as the absolute sign.

In your case  |w|=2   assuming that each of those arcs is 1 unit apart 

I would assume that that point is 

 \(2e^{(\frac{3\pi}{4}i)} \\= 2cis(\frac{3\pi}{4})\\= 2[cos(\frac{3\pi}{4})+isin(\frac{3\pi}{4})]\\ =2[-\frac{1}{\sqrt2}+i*\frac{1}{\sqrt2}]\\ =\sqrt2[-1+i]\\\)

what CPhill has written is also correct.

Here's one way:

Find the area of rectangle AEGF  = 5 * 4    =  20

From this, subtract the areas of

Triangle  ADE  = (1/2) (5)(1)  = 2.5

Triangle  CDE (1/2) (3)(1)  = 1.5

Triangle   BCG = (1/2)(4)(1)  = 2

Triangle AFB  = (1/2) ( 4)(1)  =  2

The result is the  area you want

I'll leave the computations  for you

Magnitude  of  a ±  bi =   

sqrt  [ a^2  + b^2 ] 

Find [PAB]/[PBR]. The surfaces?

\(\color{blue}A_{\Delta PAB}:A_{\Delta PBR}=2:7\)

  !

Jerry is presently twice as old as his brother and six years older than his sister. How many years from now will Jerry's age be  1/3 of the combined ages of his brother and sister at that time?

Hello Guest!

\(j=2b=s+6\\ j+x=\frac{1}{3}(b+x+s+x)\\ j+x=\frac{1}{3}(\frac{j}{2}+j-6+2x)\\ j+x=\frac{1}{3}\cdot \frac{j+2j-12+4x}{2}\\ 6j+6x=3j-12+4x\)

\(2x=-3j-12\\ -x=\frac{3}{2}j+6\)

- x |  9 12 15 18   21   24   27  30   33   36

   j |  2   4   6   8   10   12   14   16   18   20 

  s |                 2     4     6     8    10  12   14

  b |                 4     5     6     7     8     9   10 

At no point was, is and will Jerry be 1/3 the age of his siblings together.

  !

I am assuming there are only red and blue marbles in the bowl. 

Let r and b be the amount of red and blue balls in the bowl respectively. 

We have the system:

 \({r \over b} = {1 \over 3}\) 

 \({{r + 10} \over {b - 10}} = 2\)

From the first equation, we have \(3r = b\). 

Substituting this into the second equation and then solving, we find \(b = 18\), meaning \(r = 6\)

Thus, the total amount of balls in the bowl is \(18 + 6 = \color{brown}\boxed{24}\) 

The value of \(\text{boxed}(2009) = 2+4+6+8+ \cdot \cdot \cdot + 2008 \)and the value of \(\text{boxed}(2008) = 2+4+6+8+ \cdot \cdot \cdot + 2008 \)

Can you take it from here?

The triangle inequality states the sum of the side lengths of the 2 smallest sides must be greater than the length of the largest size. 

To minimize the length of the third side, we want the longest side to be 17. 

Can you take it from here?

Note that regardless of the second digit, there will always be 90 ways to choose the first and third digits (9 for the first, and 10 for the third).

If the second digit is a 0, there are 10 choices for the final digit.

If the second digit is a 2, there are 8 choices for the final digit. 

If the second digit is a 4, there are 6 choices for the final digit.

If the second digit is a 6, there are 4 choices for the final digit.

If the second digit is an 8, there are 2 choices for the final digit.

This means the number of 4 digit numbers is: \(9 \times 10 \times (10 + 8 + 6 + 4 + 2) = \color{brown}\boxed{2,700}\)

Also: Two side lengths of a triangle are 17 and 11. What is the shortest possible integer length of the third side of the triangle?

Note that triangles DXE  and  EYF  have the same height as triangle  DEF  and their combined bases = DF

So.....their total areas must be the same as  the area of  triangle  DEF

Since rectangle  DXYF  =  40......then it is composed of two equal areas .....

 So.......[ DEF  ]   =  40   / 2    =  20

1/ x^(-3/6)  =   1 / x^(-1/2)  =   x^(1/2)    =  sqrt (x)

   { Remember that    1/a^(-n)   =  a^n }

By the Triangle Inequality :

25  +  shortest side  > the remaining side

If shortest  side =  1      remaining side =  34    not possible

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If shortest side =  5     remaining side =  30      not possible

If shortest side  =   6   remaining side =  29

The shortest possible side = 6  inches

Using an exponent  law, we can write P  as

5 ^ ( 1/5 + 1/25 + 1/125  .....)

The  sum  of the exponents is the sum of an infinite series = 

(1/5)  / ( 1 - 1/5)  =  (1/5) / (4/5)   =   1/4

So P  can be expressed as

5^(1/4)

a = 5        b  =   1        c   = 4

a + b + c   =   10

C(5,1)   =   5 different amts

C(5,2)  =  10 different amts

C(5,3)  = 10 different amts

C(5,4) = 5 different amts

C(5,5)  =  1 amt

So

31  different amts

M of first earthquake =   log 10^5.1  =   5.1

M of second earthquake  = log 10^7.7   = 7.7

Ratio of M's   =    7.7  / 5.1 =   1.509     =    about 2 times as strong based on their Richter scale magnitudes

Let f(x)   = ax^2 + bx +c

g(f(x))  =   (ax^2 + bx + c)^2   - 11(ax^2 + bx + c)  + 30

g (f(x))  = a^2 x^4 + 2 a b x^3 + 2 a c x^2 + b^2 x^2 + 2 b c x + c^2 - 11ax^2 - 11bx - 11c  + 30

g(f(x)) =  a^2x^4    + 2abx^3  + (2ac + b^2 - 11a)x^2 + (2bc - 11b)x + (c^2 - 11c + 30)

And  g (f(x))  = x^4    -8x^3  + 11x^2   + 20x +  6 

Equating coefficients

a  = 1  or  -1

b = - 4 or 4

Let  a  =1   and b  =  -4

(2ac + b^2 - 11a)  = 11   →   2c + 16 - 11  = 11  →  c = 3

So

f(x)  =   x^2  - 4x  +  3

And

f(7) =   7^2  - 4(7) + 3   =     24

a/b  = 2      b/c   =  4/3

Note that   (a/b)   *  ( b/c)  =  ( ab ) / (bc)    =  a / c   =      2 * 4/3  =   8/3  

Here's what I get

A)   Assuming that the chairs  of each color are identical

You have calculated the  the number of possible arrangements correctly

If two white chairs  are together  they could occupy  positions 

1,2    and the other white chair can occupy positions  4,5,6,7 or 8

2,3    and the other white chair can occupy positions 5,6,7 or 8

3,4    and the other white chair can occupy positions 1, 6, 7 or 8

4,5    and the other white chair can occupy  positions 1,2, 7 or 8

5,6   and  the other white chair can occupy  positions  1,2,3 or 8

6,7   and the othe white chair can occupy  positions  1,2,3 or 4

7,8   and the other white chair can occupy positions 1,2,3,4,5

So.....we have   30  arrangements here

Three white chairs together  can occupy   any of 6 positions

So......the total number of excluded arragements =     36  

So....the total number of  acceptable arrangememts =   5040   -  36   =   5004

Thank you CPhill, 

The answer seems to be wrong. I think there must be an error on my side. I'll check again and plug it in WolframAlpha.

Triangles CPD  and  APB  are similar

The scale factor  between these  is   (4/2)   = 2

So....the area of triangle  APB  = [ CPD] * 2^2  =  5 * 2^2  = 20

Next...the  height of triangle   CDP  can be found as 

5 =(1/2) 2 * h  

h =  5

So....the height of triangle APB is twice this  =  10

So.....the total height of the figure =  5 + 10  =  15

And the area of triangle  CPB  = area of triangle  CDB  - area of triangle CPD  =

(1/2) (CD) * height of figure    - [ CPD]   =  (1/2)(2)(15)  -  5   =  10

And [CPB ]  = [ DPA]

So.....the total area  of the figure =  [ CPB] + [ APB ] + [ CPB ] + [ DPA]  =

5 + 20 + 10 + 10  = 

45  =    [ABCD ]