CPhill

Since  AD  bisects angle  BAC....we  have this relationship

AC / CD =  AB / BD

AC / [ 2BD ]  = AB / BD

AC / 2  =  AB

AC / 2 =  10

AC  = 20

And  we  have  this  :

area  of ABC  =  (1/2)(AB)(AC) sin BAC    ....so....

50  = (!/2) (10)(20) sin BAC

100  = 200 sin BAC    divide  both sides by 200

1/2  =  sin BAC

Using the  arcsin  we  have that

arcsin (BAC)  = 1/2 =   30°

And  since AD bisects  BAC.....then   BAD  = 15°

Irrational

sqrt  (2/9)  = 

sqrt (2)  / sqrt (9)  =

sqrt (2)  / 3  =

(1/3)sqrt (2)

An since sqrt (2)  is irrational then so is any  rational multiple   of it

OK.....here's what  I  get.......

Let x be  the  additional amount over the  original $200

The greater side of the  inequality  is

Original  pile  =  (200 + x)

After  she gives away  $50  she has   (150 + x)

And  she gives (2/3)  of  this away  =  (2/3) (150 + x)

So....she  must  have  (1/3)  of this left =  (1/3)(150 + x) = amt in her right pocket

The lesser  side of the inequality  is

(x + 200)  -  200   =   x

So  we  have  that

[150 + x ] / 3  >    x

150 + x  >  3x

150  >  2x

75 > x

x < 75 

So...I get  that  the original  amount  could be on this interval   (200 , 275)

We  can use  the  7th row  of Pascal's Triangle  to  solve  this

Number  of  Toppings         Number  of  Possible Combos

         0                                           1

         1                                           7

         2                                          21

         3                                          35

         4                                          35

         5                                          21

         6                                            7

         7                                            1

Assuming that  we can choose to have  no toppings.....the probabilty  of two toppings = 

        21 / 2^7  =   21/ 128

However...if we  must have at least  one topping the probability  of getting two toppings  = 

       21 / [ 2^7 - 1]  =    21/  127

See the answer  here  :  https://web2.0calc.com/questions/help-asap_81716

Let  x be the  liters  of the  15%  solution

So......70 - x  =  the liters of  the  35%  solution

We have this :

.15x  + .35 (70  - x)    =  .25(70)     simplify

.15x  + 24.5  - .35x  =   12.5

-.20x +  24.5  =  12.5     subtract  24.5 from both sides

-.20x  =  -12      divide both sides by  .20

x =  60  liters of the 15%  solution

70 - 60  =   10 liters of the  35% solution

THX, EP!!!

Let  x  be  the number of lbs of  the  cheaper  nuts

So....30 - x =   the  number of lbs  of the  more expensive nuts

We  have this equation....

1.10 (x)   +  2.40 (30 - x)  =   2.15 (30)     simplify

1.1x  + 72   - 2.4x =  64.50        

-1.3x  + 72  =  64.50     subtract 72 from both sides

-1.3x  =  -7.50     divide  both  sides  by  -1.3

x =  75/13   lbs  of  the  cheaper  nuts

30  - 75/13  =  315/13  lbs of the more expensive  nuts 

(2x - 7) ( x - 3)

____________   ≥  0

        x

Note  that  

2x - 7    ≥  0    

2x  ≥  7

x   ≥ 7/2

So  all  values of  x  ≥ 7/2   will  make the  inequality true

And  note  that   when    x  is on the interval    (0 , 3 ] it will also be true

And when  x  < 0    it is never true

So......the intervals that make this true  are

(0, 3 ]  U  [ 7/2, inf)

Here's  the  graph to show this  :

https://www.desmos.com/calculator/zrszwkjmby

\(f(x) = \sqrt{\dfrac{x+1}{x-1}}\quad\text{and}\quad g(x) = \dfrac{\sqrt{x+1}}{\sqrt{x-1}}\)

Note  that   f(x)  will  be defined  whenever   x  ≤  - 1     and  x > 1

But  note  that  g(x)  is  only  defined  whenever  x  > 1......

See the graph here to get a feel for this  :

https://www.desmos.com/calculator/lhfhhtc98o