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I'm not familiar with this.....I checked my answer with WolframAlpha and it syncs up....

I see  that the Shoelace Theorem is on the internet....I'll look at it, and....if I can....I'll give you the solution using it..

The rule for divisibility by 8 is that, if the last three digits are divisible by 8, then the whole number is divisible by 8

The total number of outcomes is  6^8

Notice that  the smallest that the last three digits can be  =  111

And the largest that they can be  = 666

The number of integers between these two numbers that are divisible by 8  =

floor [ (666 - 111) / 8 ]  = 69

So....the number of outcomes that are divisible by 8  =  (6)^5 * 69

So the probability  =   (6)^5 * 69  /  (6)^8   =   69 / 6^3   =   69 / 216   =  23 / 72

Actually, the answer was 3. I also got 5 initially, but the answer key is always correct. Nice try! But how do you do it?

Is there a way to do it with Shoelace Theorem?

We can use something known as "Pick's Theorem" here

The area is given by :

Number of boundary lattice points on the triangle/ 2   + number of interior lattice points   -  1

Where a lattice point is one with integer coordinates 

Number of boundary lattice points  =  5  

Number of interior lattice points = 24

So

Total area  =    5/2 + 24 - 1  =   5/2 + 23  =   2.5 + 23  =

25.5  units^2

Here's the pic :

y = 2^(5-8x)     solve for x

log y  = log 2   * (5-8x)

log y / log 2 = 5-8x

5 - logy/log2 = 8x

5/8 - 1/8 logy/log2 = x   now swithch the x and y's

5/8 - 1/8 log x / log 2 = f^-1(x)

5/8 - .41524 log x

2.There are 100 students in South High School. South High School offers only Chinese and Spanish. There are 8 more students in Chinese than in Spanish, and every student takes at least one language. If 37 students take only Spanish, then how many take both languages?

These always give me some trouble....!!!!

Let N be the number that take both

37 + N   =  number who take Spanish

37 + N + 8  = 45 + N  = number who take Chinese

So

Number ( Spanish + Chinese)  = Number(Spanish) + Number (Chinese)  - Number (Both)    = 100

(37 + N)  +  (45 + N )  - N   =  100 

37 + 45 + N = 100

82 + N  = 100

N = 18  =  number that take both  !!!!

So

37 take only Spahish , 18 take both and 45 take only Chinese  = 100

Deleted.................

5.In the Gregorian calendar, every year which is divisible by  is a leap year, except for years which are divisible by 100; those years are only leap years if they're divisible by 400.(This may seem complicated, but the calendar is carefully designed to keep the average number of days per year very close to the number of days in one complete orbit of the Earth.)Assuming we keep using the Gregorian calendar, how many leap years will there be between 2001 and 2999?

We have   998 years

floor [ 998/4 ]  = 249  possible leap years

But  the years 2200, 2300, 2500, 2600, 2700 and 2900 do not qualify because they are not evenly divisible by 400

So

249 - 6  = 

243 leap years 

10 (million) = ,5 * 1.21^(x/2)        divide both sides by .5

20 = 1.21^(x/2)       Take LOG of both sides

log 20 = x/2  Log 1.21    divide by log 1.21

15.716 = x/2      mult both sides by 2

x =31.43 years

1988 + 31.43 years = 2019

4.There are 40 animals at a petting zoo, all of which are goats or rabbits, and each animal is either an adult or a child. There are twice as many adult rabbits as adult goats, and three times as many child goats as adult goats. If there are an equal amount of adult rabbits as child rabbits, how many adult goats are there at the petting zoo?

AR = Adult rabbits

AG = Adult goats

CR = Child rabbits

CG = Child goats

2AG = AR            

3AG = CG            

AR = CR               

AR + AG + CR + CG  = 40        substitute

2AG  + AG + AR+ CG  = 40

2AG + AG + 2AG + 3AG  = 40

8AG  =  40         

AG = 5

3.If you write out the numbers from 1 through 60 (including 1 and 60), how many numbers have a "5" in them?

5 have a  "5" as a units digit  [ not including 55]

10 more  [ 50 - 59 ]  contain a "5"

So

15  contain a "5"

a) What does t = in the year 2000?    Sub that in.

b) exponential  

c) it is 2019 now.  What is t?  Sub that in.

1.Jerry writes down all the odd numbers 1, 3, 5, 7,  up to 999. How many numbers does he write down?

Let's look at something more simple....

What if he wrote down   1, 3, 5, 7 , 9   ???

Note that the number of odd integers he writes is    [ last odd + 1 ] / 2  =  [  9 + 1 ] / 2   =  10 / 2 = 5

So...this implies that the number of odds written is   [ 999 + 1 ] / 2  = 1000 / 2  =   500

Sorry...we just need to take his rate * 5 hrs  =  15 * 5  = 75 miles...!!!!

2500 = 1000 (1 + .028/12)^(12t)     divide both sides by 1000

2500/1000 = (1 + .028/12)^(12t) 

2.5  =  (1 + .028/12)^(12t)     take the og of both sides

log (2.5)= log (1 + .028/12)^(12t)     and we can write

log(2.5) = (12t) log ( 1 + .028/12)       divide both sides by 12 log (1 + .028/12)

log (2.5) / [ 12 log (1 + .028/12] = t ≈  32. 8 years   

Here's the graph :  https://www.desmos.com/calculator/iihh2mezsg

Another way:

12T = 15(T-1.25), solve for T
T = 6.25 - hours will have passed from noon.
6.25 x 12 mph = 75 miles - when Anthony catches up with her.

But its asking for the number of miles she traveled before she wwas caught up to

Note that the distance that Samantha covers BEFORE Anthony starts =

Rate * Time  =

12 * (1 + 1/4)hrs =

12 * (1.25)  =  15 miles

And each hour....the distance that he closes on Samantha =  15 - 12  =  3 miles

So....the time it will take to catch her is

Distance Samantha is ahead                       15

___________________________  =        ______   =     5 hrs

 Distance Anthony closes per hour               3

The area of the base is   W * 2W  =  2W^2

The volume of the container = Base area * height....which implies that

10 = 2W^2 * H   ⇒    H =  10 / [ 2W^2 ]  =   5 /W^2

So....the total surface area is given by.....area of the base + side area  =

2W^2 +  2(5/W^2) [ W + 2W]   =

2W^2  + (10/W^2}[ 3W]  =

2^W^2  + 30/W

So....the cost, C, to be minimized is  this :

C =  (base cost of materials)(base area) + (side cost of materials)(side area)  

C = 5(2W^2)  + 3(30/W)

C = 10W^2  + 90/W

I can do this in 2 ways

(1) Calculus.....take the derivative of the cost....set to 0  and solve

So we have

C'  = 20W  - 90/W^2  = 0

20W   = 90/W^2

W^3  = 90/20

W^3  = 9/2

W^3 = 4.5

W = (4.5)^(1/3)  ≈ 1.651

Subbing this back into the cost function, the minimized cost is

C  = 10[(4.5)^(1/3)] ^2  + 90/[4.5]^(1/3)  ≈  $81.77

(2)  If you haven't had Calculus...we can use a graph here : https://www.desmos.com/calculator/e0v6iwu5bo

Note that the function's minimized cost is ≈  $81.77

XD                   

Yeppers..... keep going......you can do it....fewer choices now available makes it easier.