CPhill

1.  Let the width  = W

The length is 6 more than  this  = W + 6

So

Area  = W * L      and the equation becomes

40  = W ( W + 6)

2. We could actually "cheat" at this point and guess the solutions rather easily...but let's do that "boring" math !!!

We have

W(W + 6  ) = 40       simplify

W^2 + 6W = 40

W^2 + 6W  - 40  = 0          we need two numbers that sum to 6 and multiply to -40....10 and -4 seem to work...so we have

(W + 10 ) (W - 4)  = 0

Set each factor to 0 and solve for  W

W + 10  =0                                                                                      W - 4  = 0

W  = -10   (reject...Width has to be positive )                                  W  = 4    this is correct

And the Length is 6 more than this  = 4 + 6  = 10  ft   

4. Experimental probability is just based on the results obtained...so...

P(Heads)  =  44 heads in  70 tosses  =  44/70  = 22/35

5.   Note...that if Teesha  is not considered for selection...the probability of selecting any other person first  is  38/39  and the probability of selecting a second person ( but not Teesha)  is   37/38

So....the total probability  is      38/39 * 37/38   =   (38/38) * (37/39)  =  1 * 37/39  = 37/39

P......the student should have divided by 9 from step 1 to step 2

1.  7!  ways  = 5040   ways

2. P (8, 5 )  = 6720

3. If 16 were green, 64 - 16 =  48 must be some other color....so...the probability of  non-green  is

48 / 64  =  3 / 4  =  75%

x^2  - y^2  = (x + y) (x - y)

We need   to have   (52)(48)

It's obvious that x must be < 52....because,if not, ...... y is positive.... and adding them together would produce  more than 52

So x must either be  50 or 48

Notice that if x  = 50 and y  = 2, we have

(50 + 2)  (50 - 2)  =

(52) (48)   !!!!

If we "anchor" any one person in a seat, there are  8! arrangements of the other peoeple that are possible = 40320  arrangements

If we "anchor"  the mathematicians in any 4 consecutive positions there are 4! = 24 arrangements of them that are possible

And the other people can be arranged  in 5!  = 120 ways

So....the total number of ways that the mathematicians can be seated is  24 * 120  = 2880  ways

So....the probability that all 4 math majors sit in consecutive seats  is      2880 / 40320  = 1/14 ≈  7.14%

Thanks, Rarinstraw  !!!

Here's one way to do it, ACG..but maybe not the best...sorry I can't provide a pic...!!

Draw angle bisectors to A, B and C

Where these bisectors meet is the incenter of the triangle....and this will  be the center of the circle....call this point, O

Now...looking at the circle, FE, FD and DE  form chords within the circle....and any chord passing through O will be a diameter....but none of these chords will pass through O....and any chord in a circle not passing through the center will intercept an arc on the circle that is  less than 180°

Thus....inscribed angles FDE, EFD and and FED will intercept arcs that will be less than 180°....so each of their measures will be less than  (1/2)  of 180°  =   less than 90°

So...triangle DEF is acute because each of its angles measure < 90°

I hope that helps ....!!!

Note.....   ⁿ√a^m   =  a^m/n  .... so.....

⁵√w³  is equivalent to    w^3/5  =  ( w^1/5 ) ³  ⇒​   "D"

Second one

Let the width be, W

So the length is 10 more than this  = W + 10

And 

Area  = W * L  =  W (W + 10)  = f(W)

Volume  =

(1/3) 3.14 * (30/2)^2 * 60  =

20 * 3.14 * 15^2  =

20 * 3.14 * 225  =

14130   ft^3