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I love the game set. :))

(a) How many cards are in a complete deck of Set?
3^4

(b) How many unique sets are there?

81*80/3/2 = 1080

(c) Find the number of sets where all three cards are the same for exactly 0 attributes.
Hm, this is where we seem to disagree. 

I think you calculated the total number of 3 cards that wouldn't make a set. 

If you are unfamilar with the game, a set can be made where all the attributes are all the same or all different. 

For example, (1, red, oval hollow) (1, purple, diamond shaded) (1, green squiggle solid) would make a set because all the attributes are all different or all the same. 

Nice job for question a and b. :))

Your solution with b was really clever. 

=^._.^=

A = 1/2[3(3 * sin45º)]

Thank you!!!!!!!!!!!! That is correct!!!!!

https://web2.0calc.com/questions/geometry_10279#r2

I am afraid that is wrong. I don't know why yet, but I don't think you do \(\dfrac{115}{2}\).

Arc WY - Arc XV = Angle U = 30 degrees.

Arc WY + Arx XV = 360 - 170 - 110 = 80 degrees.

Therefore, arc WY = (30 + 80)/2 = 55 degrees.

AC = 8

∠CAE = 45º

∠C = arctan(6/8)

∠AEC = 180 - (∠CAE + ∠C)

CE = (sinCAE * 8) / sinAEC

EF = sinC * CE

You can do the rest. 

Both of the separate triangles in this figure are isosceles, as two of their sides are the same. 

Thus, we know that $\angle{ACD} = \angle{DAC} = 25$. 

From this, we know that $\angle{ADC} = 180 - 25 \cdot 2$, for the two other angles in the triangle. Thus, $\angle{ADC} = 130$. 

Because $\angle{ADB}$ and $\angle{ADC}$ are supplementary, we know that $\angle{ADB} = 180 - 130 = 50$. 

Finally, since $\triangle{ABD}$ is isosceles, we know that $\angle{ABC}$ is also $\boxed{50}$. 

Angle ACB  =  180  -  60  - 45   =  75

And if AT  bisects  CAB  ....then angles CAT  and  BAT  =  30

Using  the Law of Sines

CT  =  24 sin (30°) / sin (75°)  =    12/sin (75°)   = 12csc(75°)

TB  =  24 sin (30°)  / sin (45°)  =   12 / sin (45°)   =  12csc (45°)

So    CB   =   CT  +  TB   =     12   ( csc (75°)  +  csc (45°) )    ≈  29.39   

See here:

https://web2.0calc.com/questions/in-isosceles-right-triangle-abc-point-d-is-on-hypotenuse

Because the triangle is isosceles we know that $\angle{ABC} = 45$ ($\frac{180-90}{2}$)

Because it trisects angle $ABC$, we know that $\angle{EBA} = 15$. Thus, we subtract $180 - 15 - 90 = 75$, because $EAB$ is a triangle. 

Therefore, our answer is $\boxed{75}$. 

Mmmmm.....haven't done one like this before

A  graph  seems the  best  way to go, here :  https://www.desmos.com/calculator/w1pkainlsv

The  solution point   is  (x , y)   =   (8 , -2)

p(q(x))  means we  are putting  q  into  p ....so....

(2x^3  + 1)^2   =    4x^6  +  4x^3   + 1

Adding   this to r(x)   gives  us

7x^4  + 4x^6  - x^3   - 6x  - 1

If a, b , c  are consecutive  integers   a  =  2    c  =  4    and b   = 3

So....the  area  =   (1/2) (a + c) (b)  = (1/2)(6)(3)    =   9  units^2

Wow, this makes sense now! Thank you so much @catmg :))

1. A teacher arranged her books on a shelf. She has 4 astronomy books, 3 Biology books, and 2 Chemistry books. How many ways can she arrange these books on the shelf if the books from each subject must be together?

The  number of ways  =    arrangements in each topic * arrangements of topics =   4!  * 3!  * 2!  *  3!  = 

1728  arrangements

2. How many ways can David, Sam, Emma, and Lanie stand in a line if David and Sam refuse to stand next to each other?

Number of   ways  without restrictions  =  4!  =   24

Number  of ways  if David and Sam DO  stand next to each other   =  3 places they can occupy * 2 arrangements in each place   =   6

So.....number of  ways  if  they refuse to stand next to each other  =  24  -  6   =   18 ways

(2x  + 1/2)^6

We  are looking  for   the  third  term

This  is  given by

C(6, 2)  (2x)^4  (1/2)^2  =

15 *  ( 16x^4)  ( 1/4)  =

15 *  4x^4  =

60 x^4

If BAC  =  23

Then   QAD  =  90 - 23   =  67

And QDA  =  23

So

QDA  +  QAD  =    23  +  67   =   90  (degrees)

I'm assuming the question equation is 

$$(2x+1)^6$$

in which the answer is $240$ by the binomial theorem, 

https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:polynomials/x9e81a4f98389efdf:binomial/v/binomial-theorem

Note that if DE = 3EC

Then    3EC  + EC  =  20

4EC =  20

EC =  20 / 4    =  5

So  DE =  3 * 5  =  15

Note that riangles DFE  and  BFA  are similar 

And  DE / BA   =    15/20  = 3/4

Which also means  that    DF / BF  =   3/4

So    BF / BD    =  BF / (BF + FD) =  4  / (4 + 3)  =     4/7

And  triangle  FGB is similar to triangle DCB

So

FG / DC     =     BF / BD

FG /  20  =   (4/7)

FG  =  20 (4/7)   =   80/7

2. 

a. 10*9*8=720

b. 10*10*10

Oops i got my second question wrong sorry

26 red cards +  6 black ( J - Q -K)

32 out of 52 = 8/13

36 possible rolls

  the rolls that are less than 4

     11

      12

        21         3 out of 36      1/12

1. There are 52 cards in a standard deck. There are 4 jacks, 4 Queens, and 4 Kings. In addition to this, there are 26 red cards

(4+4+4)/2= 6 (because 2 out of the 4 jacks, queens and kings are red)

6+26=32

32/52 is the probability.

You can solve it from here

2.Total possible number of combinations

6*6=36

The number has to be less than 4,

so

1+1

1+2

2+1

1+1

4 combinations 

4/36=1/9

So the probability is 1/9

Each compounding PERIOD is 6 months      2 per year

   interest per period in decimal form is   .10/2 = .05

    12 months is two periods

        10 000 ( 1 + .05)² = amount owed after 1 year = __________  dollars