CPhill

(2, 0)     (0, 2)

Te slope of this line is      [ 2 - 0 ] / [ 0 -  - 2 ]  =   2 / 2   =   1

And the equation of the line is

y = 1 ( x + 2)

y  = 1x + 2

So....m + b  =    1  + 2   =   3

We have this

( 4  + x) / 2  =  2                (3 + y) / 2  = 9

 4 + x  =  4                          3 + y  =  18

x = 0                                   y =  15

So...the other endpoint is  (0,15)  and the sum of its coordinates is  15

p(t) = -16t^2 + 800t - 4000

a) The price of the ticket that maximizes the profit  is given  by  :

-800 / ( 2 * -16 )  =   -800 / (-32)  =  $25

b) The max profit  is    -16(25)^2 + 800(25) - 4000  = $6000

c)  5424 = -16t^2 + 800t - 4000    subtract 5424 from each side

-16t^2 + 800t - 9424  = 0       multiply through by -1

16t^2 - 800t + 9424 = 0   divide through  by 16

t^2 - 50t + 589  = 0    factor as

(t - 31) (t - 19)  =  0

Setting both factors to 0 and solving for t produces two possible prices...$19 or $31

2/17  =  1/a + 1/b

Assuming that a, b  are different integers

We can use something known as "Egyptian Fractions"  here 

Divide  17/2  =  8.5     .....take the next greatest integer  = 9   and write  1 over this  = 1/9

Subtract this from 2/17

2/17  -  1/9  =  1/153

So

1/9  + 1/153  =  2/17

So  { a, b}   =  ( 9, 153} or {153, 9 }

Not totally sure about this, but.....

If a : b  is  5 : 6  there are 11 equal parts

If b : c  is 8 : 9  there are 17 equal parts

And the common denominator is  11 * 17  = 187

So  a possible value for  b  = (6/11) * 187  =  102

But c  =  (9/8)b  =  102 * 9/8  =  114.75

But c needs to be an integer

So...mutiplying this by 4 will give us the integer 459

So....multiplying b by 4   we get 102 * 4 =  408

And  a  is  (5/6)b  =  (5/6)(408)  =  340

So

a , b , c  =    340, 408, 459

And   a: b   =  340: 408   =  5/6     And  b : c  =  408: 459

So....a + b + c  =     340 + 408 + 459   =   1207

Sorry, hectictar....I didn't see your answer when I posted....!!!

I get  99

9 + 9  =  18

99 / 18   =  5.5

If PQ  is 6 times the length of QR, then  PR  can be divided  into (6 + 1)  =  7 equal parts

So....each equal part  is   98 / 7  =   14 cm

So..... QR  is  one of these equal parts  = 14 cm

And PQ is six of the equal parts  =  6 * 14   =   84 cm

1. A board USB cut into 20 equal pieces so that each piece is 6 1/2 cm in length. How long was the board before it was cut?

20  *  (6 + 1/2)   =   20*6  + 20* (1/2)  =  120 + 10  =  130 cm

2. Mr. Phillips served 1/2 of a chicken for dinner. 2/3 of the chicken served was eaten. What fraction of the roast served was eaten?

Well.....this could be construed as a" trick " question...if 2/3 of what was served was eaten, then 2/3 must be the answer

However...if we are talking about the fraction of the total roast eaten, we have

(1/2) (2/3)  =  2/6  =   1/3  of the whole roast was eaten

3. Mario completed 3/5 of his homework. One fourth of the completed homework was in geography. What franctional part of the homework was in geography?

(3/5) (1/4)  =  3/20   

The scale factor  from   ABC  to RQR  is      6/4  =  3/2  =  1.5

So....each dimension  of  PQR  is 1.5 times that of ABC

Now....just for the heck of it,if we let the base of ABC  = 4 cm...so.... its altitude must be 10 cm

Proof  ..area ABC  =   (1/2) base * altitude  =  (1/2) (4) (10)  =  20  cm^2

So....since each dimension of PQR  is 1.5 times that of ABC, its altitude must be 10 * 1.5  = 15 cm

And let its base  = 6cm

So...it's area is just  (1/2) (6) ( 15)  =  45 cm^2

Note that  we could obtain the same result  by taking the area of ABC  and multiplyng this by the square of the scale factor

Proof :   20 (1.5)^2   =  20 (3/2)^2  =  20 * 9 / 4  =  180  / 4   = 45  cm^2

Hope that helps  !!!