CPhill

Note that   Distance / Rate  =  Time

Since they travel the same time......let the distance that A drives  = D

And let the distance that B drives  = 180 - D

So....equating times [ in hours ], we have that

D / 60  =  [ 180 - D ] / 48    cross-multiply

48D  = 60 [ 180 - D ]     simplify

48D  = 10800 - 60D      add 60D to both sides

108D  =  10800       divide both sides by 108

100 = D  ....so ...the time they meet  =  [D / 60 ]  hrs.  =  [ 100/ 60 ]  hrs.  =  [5/3]  hrs   =

 [5/3] * 60 minutes =   100 minutes after 8AM  =  9:40 AM

Here's another approach using the Law of Cosines only once :

Refer to the following image :

RT^2  =  ST^2 + RS^2  -  2(ST * RS) cos (RST)

15^2  =  14^2 + 13^2  - 2(14 * 13) cos (RST)   simpllfy

225 - 196 - 169  = -364 cos(RST)

-140 = -364cos(RST)

cos RST  = 140 / 364  =  5/13

Now....in triangle RMS draw altitude RN and note that angle RSN  = angle RST........since triangle RNS is a right triangle with hypotenuse RS = 13......then.......if  cos RST = 5/13...it must be that  SN = 5.......so triangle RNS  has a hypotenuse of 13 and one leg   = 5.....so...the other leg (RN)  must = 12

And triangle RMN is another right triangle with RN  = 12.....and if M is the midpoint of ST, then SM  = 7.......but SN was shown to be = 5....so.....MN  = 2

And we have that  MN^2 + RN^2  = RM^2   ....so.....

2^2  + 12^2   = RM^2

148   = RM^2

√148  =  RM

6 sin^2(x) * cosx - 3cosx  =  -3 cos (x) cos (2x)    

We can work on both sides  and show that they are equal.....

First, factor out cos x  on both sides  and write cos 2x as 1 - 2sin^2 x 

cos x [ 6sin^2 x  - 3 ]  =  cos x [ -3 (1 - 2sin^2 x) ]

Simplify

cos x [ 6sin^2 x - 3 ] = cos x [ 6sin^2 x  - 3 ]

We  can use the Law of Cosines twice here

RT^2  =  ST^2 + RS^2  -  2(ST * RS) cos (RST)

15^2  =  14^2 + 13^2  - 2(14 * 13) cos (RST)   simpllfy

225 - 196 - 169  = -364 cos(RST)

-140 = -364cos(RST)

cos RST  = 140 / 364  =  5/13

And

RM^2  = SM^2 + RS^2  - 2(SM * RS) cos RST

RM^2  = 7^2 + 13^2 - 2 ( 7 * 13) (5/13)  simplify

RM^2 =  49 + 169 -  [ 2 *7 * 5 ]

RM^2  =  148

RM  = √148  =  √ [ 37 * 4 ]   =  2√37 ≈  12.166

1)Tim Worker expects Social Security benefits of $937 per month. His wife, Mary, is eligible for a wife's benefit of 50 percent of Tim's benefit. How much will Mary receive?

50%  of  937  =  .50 * 937  = $468.50

2) The Brown family's dinner bill was 75.89 and they left 10.00 as a tip. What percent was the tip?

Assuming that 75.89 was the total bill.....the percent of the tip is given  by :

10.00 / 75.89  ≈ 13.2 %

As for the third question....I don't know.....!!!

LOL!!!!!!......we'll see......don't get too cocky....!!!!

I think we've worn this one out.....

By the order of operations......multiplication and  division arre carried out at the same level of precedence......that is....we evalutate them from left to right as we encounter them....so...

6/2 (2 + 1) .......we encounter the division of 6/2, first  ...and this =  3

3 ( 2 + 1)      simplifying inside the parentheses

3 (3)  = 9

y = e^x cos (x)

y '  = e^x cos(x)  - e^x sin(x)

y "  = e^x cos (x)  - e^x sin (x)  - e^x sin (x)  - e^x cos(x)  =  -2 e^x sin (x)

So

y"   -  2y' + 2y  =

 -2 e^x sin (x)  - 2 [  e^x cos(x)  - e^x sin(x) ] + 2 [ e^x cos(x) ]   =  0

f " (x)  = 2x   f ' (1)   = 2, f (0)  = -1

f "(x)   = 2x        integrate to find the first derivative

f ' (x) =  x^2  + C₁      applying the initial condition to find C₁ ,  we have that

f '( 1)  = 2

2  =    (1)^2 + C₁

2  = 1 + C₁

1  = C₁

So....  f'(x) =  x^2 + 1

Integrate this again to find y(x)

f(x)  = (1/3)x^2 + x  + C₂

And applying  the  second initial condition to find C₂, we have that

f(0) = -1

-1 = (1/3)(0)^3 + 0 + C₂

C₂  = -1

So.....   f(x)  =  (1/3)x^3 +  x   - 1

x + y = 9      →       x  =  9 - y      (1)       note......this isolates x

-10x + 6y = 6     (2)

Sub (1)  into (2)

-10 ( 9 - y) + 6y  = 6     simplify

-90 + 10y + 6y  = 6

-90 + 16y  = 6      add 90 to both sides

16y = 96     divide both sides by  16

y  = 6          and using (1),    x =  9 - 6   =   3