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Could you retype your question so it is clearer? What is it that you are looking for?

- x + y + z = 30

- 5 + 12 + 25 = 32

Your answer is not correct!

x + 3y + 2z   = 1        (1)

-3x+ y + 5z   =10       (2)

-2x -3y + z   = 7         (3)

Adding (1)  and (3)   we get

-x +3z   = 8      (4)

Multiply  (2)  by  3  =   -9x + 3y  + 15z    = 30

Add this to    (3)    and we  get

-11x + 16z  = 37    (5)

Multiply   (4)  by -11   and we  get      11x  - 33z = -88

Add this  to  (5)  and we  get that

-17z  = - 51

z = -51/-17   =  3

Sub yhis into (4)  to find  x

-x + 3(3)  = 8

9 - 8  = x

1   = x 

And using (1) to find y

1 + 3y + 2(3)  = 1

3y + 6  =0

3y  =  -6

y = -6/3

y  = -2

{ x, y, z }  =   { 1, -2, 3 }

Two sectors of a circle of radius 12 overlap as shown, with P and R as the centers of the respective circles. Determine the area of the shaded region.

Shaded area = 2*[(r²pi / 4) - r² / 2]

Thank you so much CPhill! :)

Thank you so much CPhill! 😊

Thank you!

201 INCLUDES 100 and 300

199 is the number of numbers BETWEEN 100 and 300 as asked

x^2  + y^2   = 2x  + y

x^2 - 2x  + y^2  - y   =   0      complete the square on  x and y

x^2  - 2x + 1  + y^2  - y  + 1/4  =   1 + 1/4      factor  

(x - 1)^2  +  ( y - 1/2)^2  =   5/4

This is a circle  with a center of   (1, 1/2)   and a radius of  sqrt  (5/4)  =  sqrt (5)  / 2

So.....the largest x that will provide a real numer for y  is   when  x =  (1 + sqrt (5)/2) 

Using the law of cosines we have:

4² = 5² + 6² - 2 * 5 * 6 * cosA

cosA = (5² + 6² - 4²) / 2 * 5 * 6   ==>     cosA = 3/4

The folllowing pic  shows something interesting :

Any  such line with the form    y = mx +  b      where  b = 1   will always  divide the square into congruent quadrilaterals

So....the y intercept =  1

there is actually a formual for doing these types of problems. For finding the numbers of numbers between 2 such numbers a and b, the answer is             (b - a)+1. Using this, you can plug in the 2 values and get 

300 - 100 = 200

200 + 1 = 201

201 is your answer

using a graphing calculator (I used desmos) you can insert the 2 equations and get an intersection with 2 points (.94, 4.911), and (-4.171, 2.757). Using these 2 points, use the distance formula ( d=√((x_2-x_1)²+(y_2-y_1)²) ). Plug in the values and you will get √30.762037. 

199 is the answer....

The question is on the header

What is the question, dududsu2  .    ????

∑[ (-1)^(n+1) * n, n, 1, 2010] =-1005

-1005 mod 2010 = - 1005

2010 + (- 1005) = 1005

Let  x =          1

                 ______

                          1

                 2 -  _____

                        2 -   1

                             _____

                              2 - .....

So we  have

x  =      1

      ________               multiply   both sides by  2 - x

        2  -   x

x (2 - x)  =  1    

2x  - x^2  = 1      rearrange as

x^2    - 2x   +   1  =  0        factor

(x - 1) ( x - 1)  =  0

(x - 1)^2   = 0        take both roots

x - 1    = 0

x   =1

Use the law of sines to find AC, then use the same law to find the length of side AB.

299 is the answer.

From the graphs and the data, you can see that anna and elsa BOTH had LONGER jumps than olaf, but the AVERAGE for olaf is longer.

   it depends on how the meet is scored to determine the winner, I think.....  do the jumpers take the average of several jumps (olaf wins-send him) ...or do they take the LONGEST of several jumps?  anna (send her) beats elsa beats olaf ....