CPhill

The sum is given by:

[19 + 10] * 5  =  29 * 5 = 145

Here's 2

The lateral surface area is given by:

A = pi(R + r)L      where R, r are the radiuses of the larger and smaller bases, respectively, and L is the length of the lateral side  ......    and L =  sqrt(h^2 + (R - r)^2)    so we have

A = pi(9 + 4)sqrt(12^2 + (9 - 4)^2 )  =  13pi*sqrt(144 + 25)  =  13pi* sqrt(169)  =  13* pi * 13  =

169pi  cm^2

Here's 1......

V = (pi * h)/3   * (R^2 + Rr + r^2)      and we have

224pi^3 = [(pi * 6)/3] [R^2 + R(12-R) + (12- R)^2]

224 pi^2 = 2 [ R^2  + 12R - R^2 + 144 - 24R + R^2]

224pi^2 = 2 [R^2 - 12R + 144]

112pi^2 = [R^2 - 12R + 144]   .....solving this equation for R we get that R ≈  37.582

This is clearly impossible.....but.....I solved it based on the way you have written it........did you make a mistake somewhere???

[ Is 224pi^3    supposed to be something else???]

p-4/p = 16/5p - 1/5    [I'm solving this as you have written it, literally.......I'm not sure this is what you meant, though.....!!!!!]

p - 4/p  = (16/5)p - 1/5        multiply through by 5

5p - 20/p = 16p - 1          mutiply through by p

5p^2 - 20  = 16p^2 - p      rearrange

11p^2 - p + 20 = 0    

This has no real solutions..........using the Quadratic Formula, the solutions are :

[ 1 +/- sqrt(879) * i ] / 22

Using a "weighted" average, we have

[3(1) + 4(0) +  3(2) + 1(0) ] / [11] = .818  ≈ .82 [rounded to hundredths]

10^(x-5)=5   take the log of both sides

log 10^(x - 5)  = log 5   and we can write

(x - 5) log 10   = log 5       and log 10 = 1, so we can ignore this part.....and we have.....

x - 5 = log 5       add 5 to both sides

x = 5 +  log 5  ≈  5.69897

200 + 12x  < 100 + 15x

100 < 3x

100/3 < x

33.3 < x   or

x > 33.3 people =  34 people   [ since we can't have "partial" people.....!!! ]

Let W be the original width and L be the original length

After folding we have  the new  width = (1/2)L  and the new length = W

So

L /W = W / (1/2)L   cross-multiply

(1/2)L^2  = W^2      take the square root of both sides

sqrt(1/2)L = W

sqrt(2)/2   = W / L

I don't see any diagram, Mellie.....so I'm just guessing at the answer, here.....!!!!!

Let the distance around the track = d

Then, every second, Andy runs d/40th of a lap.

Let them start at the same place running in opposite directions.

So, in 15 seconds, Andy runs  (d/40)(15)   = (15/40)d  =  (3/8)d

And since they meet, Bob must run  d - (3/8)d = (5/8)d    every 15 seconds........so........

15/ [(5/8)d] =   x / [1 d]    where x is the total number of seconds it takes Bob to complete 1 lap......

  cross-multiply.....

15d = x(5/8)d

15  = (5/8)x

(15)(8/5)  = x =   24 seconds  = Bob's time to run one lap.....

We have

[d - (1/3)d ]   = 20       and

[r + (1/3)d]  = 20

Add the equations and we get that

d - r =   40      →  d =  40 + r      substituting this into the second equation, we have

r + (1/3)(40 - r)  = 20

(2/3)r + 40/3  = 20       multiply through by 3

2r + 40   = 60        divide through by 2

r + 20 = 30         r = 10         and d = 40 - 10 = 30

So....Ramon started with 10 pieces of candy and Dave started with 30