geno3141

When  x = 0,  y = 8    --->   the height of the triangular region is 8.

When  y = 0,  x = 4    --->   the width of the trianglular region is 4.

Use the formula for the area of a triangle:  Area  =  ½ · width · height  to get the area of the region.

[  5 / sqrt(80)  +  sqrt(845) / 9  ]   /   sqrt(5)

=  [  5 / sqrt(80)  +  sqrt(845) / 9  ]   x   [ 1 /  sqrt(5) ]

=  [  5 / sqrt(80) ]  x  [ 1 /  sqrt(5) ]    +    [ sqrt(845) / 9  ]  x  [ 1 /  sqrt(5) ]

=  [  5 / sqrt(400)  ]    +    [  sqrt(169) / 9  ]

=         5 / 20             +            13 / 9                     =     ...

At first, Tom's savings  =  x     and Sean's savings  =  x.

After Tom gave Sean  1/4  of his savings, Tom had  x - 1/4x  =  3/4x

                                                         and Sean had  x + 1/4x  =  5/4x

At this time, Sean had 52 more than Tom:  Tom + 52  =  Sean

                              --->                                 3/4x + 52  =  5/4x

                                                                                52  =  2/4x     --->     x  =  104  (to start with)

Now, Tom has  3/4(104)  =  78     and     Sean has  5/4(104)  =  130     for a total of  208.

Let  x  =  the number of diners:     604 + 5x  =  988 + 3x

Solve the above equation for x; this will be the number of diners.

How much?  Use the above answer in either the expression   604 + 5x   or   the expression   988 + 3x

The number of ways that 2 female students can be chosen from 7 female students is:  ₇C₂   

The number of ways that 3 male students can be chosen from 5 males students is:  ₅C₃   

The number of ways that 2 female students can be chosen from 7 females students and 3 male students

can be chosen from 5 male students is:  ₇C₂  x  ₅C₃   

The number of ways that 5 students can be chosen from 12 students is:  ₁₂C₅  

The probability will be:  ₇C₂  x  ₅C₃  / ₁₂C₅  

Notice that CD falls outside the triangle; BA + AD = BD.

In triangle(BDC), BD2 + CD2  =  BC2                          In triangle(ADC), AD2 + CD2  =  AC2  

      --->     (BA + AD)2 + CD2  =  BC2                                                      AD2 + CD2  =  82  

                 (17 + AD)2 + CD2  =  202                                                       AD2 + CD2  =  64  

    289 + 34AD + AD2 + CD2  =  400

              34AD + AD2 + CD2  =  111 

Combining these two:          34AD + AD2 + CD2  =  111 

                                                         AD2 + CD2   =   64  

Subtracting:                         34AD                        =   47               --->     AD  =  47/34  

In triangle(ADC), AD2 + CD2  =  AC2  

                       (47/34)2 + CD  =  82

                                         CD  =  7.8796632...

Area of triangle(ACD)  =  ½·AD·DC  =  ... 

Multiplying each term by  sqrt(2) / sqrt(2)

1/sqrt(2)  +  3/sqrt(8)  + 6/sqrt(32)

=  sqrt(2) / sqrt(4)  +  3sqrt(2) / sqrt(16)  +  6sqrt(2)/sqrt(64)

=  sqrt(2) / 2  +  3sqrt(2) / 4  +  6sqrt(2) / 8                     

=  2sqrt(2) / 4  +  3sqrt(2) / 4  + 3sqrt(2) / 4               (writing each fraction with a denominator of 4)

=  8sqrt(2) / 4

=  2sqrt(2)

What about  1 + 1 + 49 + 49 + 50  ?

The first equation is:  3a + 4b  =  7

The second equation is:    6a + 4b  =  k - 3a - 8b

which becomes:              9a + 12b  =  k

Multiplying the first equation by 3:  9a + 12b  =  21

For these to have infinitely many solutions, these must be two forms for the same equation; so  k = 21

Two points on this line are  (0,-3)  and  (1,-5).

The solpe is  (-5 - -3) / (1 - 0)  =  -2

The y-intercept is  (0, -3)

The equation is:  y  =  -2x - 3