CPhill

Let total amount = T

So

Dan originally has (13/18)T   and Paul has  (5/18)T

So....after Dan gives Paul 32  pounds....we have that

(13/18)T - 32 = (5/18)T + 32    simplify

(13/18)T - (5/18)T = 64

(8/18)T =  64

(4/9)T = 64

T = 64 (9/4)  =  144

So... Paul originally had  (5/18)(144) =  40 pounds

Yours ( and mine)  are correct!!!!

Actually, asdf.....we both are correct....we just need to "fuse" our final answers!!!

I had forgotten that the triangle could also lie "below" the x axis

x               x^2 - 6x + 8           x - 2

0                     8                      -2

1                     3                      -1

2                     0                       0

So....x = 2 is the solution

The line intersects the x axis at x = -7.5

The base length of the triangle will be   k - (-7.5) =  k + 7.5

The height of the triangle will be    (2/3)k +  5

So...we are trying to solve this

(1/2) base * height = 20

(1/2) ( k + 7.5) [ (2/3)k + 5]  = 20

(k + 7.5) (2/3k + 5)  =  40

(2/3)k^2 + 5k + 5k + 37.5 = 40

(2/3)k^2 + 10k - 2.5 = 0             multiply through by 3

2k^2 + 30k - 7.5   =  0           multiply through by 10

20k^2 + 300k - 75  = 0

4k^2 - 60k - 15 = 0

The soutions to this are   k ≈ -15.246      and k ≈ .245   

So....the integer values for k that make this triangle have an area < 20    are  the integers  -15 to 0 inclusive

And the sum of these =  - ( 15) (16) / 2  =  -120 

We know that Eli as already traveled 4 km   when Philip starts

And Eli has taken 4/3 hrs to do this

So....let  the time it takes Philip to reach the halway point = T   (in hrs)

So....the distance Philip travels = 7T 

And the distance that Eli travels  = the 4 km he has traveled plus his rate times the time it takes Philip to reach the halfway point =  4 km + 3T

So....the distance that Eli travels = the distance that Philip travels  ...so we have

4 + 3T = 7T

4 = 4T

T = 1  hr

So....Philip travels 7(1) = 7 km

And Eli travels  4 + 3(1)  = 7 km

So...each travels 7km

It takes Eli (4/3)hrs + 1 hr =  7/3 hrs  to travel halfway

And it takes Philip just 1 hour to travel halfway

Here's a pic :

Since MCFN is a rhombus....then FN is parallel to CM

But FN is also parallel to DC

And CE is a transversal cutting parallel segments which makes coresponding angles EFN and ECD equal

And angle FEN = angle CED

So....by AA congruency, triangle NFE is similar to triangle DCE

Then    FN / FE  = CD /CE

And FN = CF  = 8 

And FE = 12

And CE = CF + EF =  20

Therefore

8 /  12 =  CD / 20

2 / 3 = CD /20

CD = 40/3

Therefore DE  =  55 - CE - CD =  55 -  20 - 40/3  =  65/3

An angle bisector AC divides a trapezoid ABCD into two similar triangles △ABC and △ACD. Find the perimeter of this trapezoid if the leg AB=9 cm and the leg CD=12 cm.

Let AC bisect A

Then angle CAD = angle CAB

And angle ACD = angle BAC

So...triangle ACD is similar to  triangle CAB

So

CD/AB = AD/BC

12/9 = AD/BC

But angle ACD = angle CAB = angle CAD = angle ACB

So AD = CD

And BC = AB

So AD = 12

And BC = 9

So....the perimeter of the trapezoid is AB + BC + CD + AD = 9 + 9 + 12 + 12 =  42

(n - 6) + (n - 4) + (n - 2) + n + (n + 2) + (n + 4) + (n + 6)

7n =  182

n = 26

The smallest number is    26 - 6   =   20

12 and 21 aren't primes

Number of minutes in a week =

60 x 24 x 7   = 

( 5 * 3 * 2^2) * ( 2^3 * 3) * (7)  =   2^5 * 3^2 * 5 * 7

Greatest prime factor =  7