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If you ignore wall friction , you will need TWO tubes that total the same AREA as the 5" tube

Area of 5" tube = pi r^2  = pi (2.5)^2 = 19.63 sq in

Two tubes, each would have to equal  19.63/2  sq in = 9.82 sq in

pi r^2 = 9.82    results in r =  1.768 inches.....   2 r = diameter = 3.54 inch diameter each

'i'  is DEFINED as the sqrt of (-1)      so example:    sqrt(-4)  = sqrt (4 x (-1) )  =  2 sqrt(-1) = 2 i

Talk to an Accountant!!.

Solve for y:
-2 = 2 y^2-4 y-5

-2 = 2 y^2-4 y-5 is equivalent to 2 y^2-4 y-5 = -2:
2 y^2-4 y-5 = -2

Divide both sides by 2:
y^2-2 y-5/2 = -1

Add 5/2 to both sides:
y^2-2 y = 3/2

Add 1 to both sides:
y^2-2 y+1 = 5/2

Write the left hand side as a square:
(y-1)^2 = 5/2

Take the square root of both sides:
y-1 = sqrt(5/2) or y-1 = -sqrt(5/2)

Add 1 to both sides:
y = 1+sqrt(5/2) or y-1 = -sqrt(5/2)

Add 1 to both sides:
Answer: |y = 1+sqrt(5/2)    or    y = 1-sqrt(5/2)

but why greater than in the last equation?

which angle? there are three

If a function is y=-(1-x)²-10, the parent function is y=x². If the function is y=x³-3x²+3x-1, the parent is y=x³. The parent function is the function without adding, subtracting, multiplying, dividing, or negatives.

Solve for x:
5 x^4-2 x^3+3 x^2 = 0

Factor x^2 from the left hand side:
x^2 (5 x^2-2 x+3) = 0

Split into two equations:
x^2 = 0 or 5 x^2-2 x+3 = 0

Take the square root of both sides:
x = 0 or 5 x^2-2 x+3 = 0

Divide both sides by 5:
x = 0 or x^2-(2 x)/5+3/5 = 0

Subtract 3/5 from both sides:
x = 0 or x^2-(2 x)/5 = -3/5

Add 1/25 to both sides:
x = 0 or x^2-(2 x)/5+1/25 = -14/25

Write the left hand side as a square:
x = 0 or (x-1/5)^2 = -14/25

Take the square root of both sides:
x = 0 or x-1/5 = (i sqrt(14))/5 or x-1/5 = 1/5 (-i) sqrt(14)

Add 1/5 to both sides:
x = 0 or x = 1/5+(i sqrt(14))/5 or x-1/5 = -(i sqrt(14))/5

Add 1/5 to both sides:
Answer: |x = 0    or    x = 1/5+(i sqrt(14))/5    or    x = 1/5-(i sqrt(14))/5

When their in pre-school or Kindergradin 

Omi67

I have a different way of solving

ElectricPavlov: See my explanation here:

http://web2.0calc.com/questions/m-t-85e-t-409-describes-the-amount-of-a-radioactive-isotope-as-a-function-of-time-t-years-after-how-long-the-amount-has-dropped-to-1-6-of-the

Also, please read this so that you can understand how radioactive materials decay with time:

http://math.usask.ca/emr/examples/expdeceg.html

......and you misspelled   'were'       

If there WERE two chickens....it depends into how many PIECES you cut the chickens.

Simplify the following:
(13×18)/(15-9)

(13×18)/(15-9) = (13×18)/(15-9):
(13×18)/(15-9)

15-9 = 6:
(13×18)/6

18/6 = (6×3)/6 = 3:
13×3

13×3 = 39:
Answer: |39

Not sure what you are doing on this problem Guest #2...if I substitute your answer of 96.34 into the equation I get

m(t) = 3.0666   which is NOT   1/10 of the original 85 at t=0.

I come up with 66.77 years like Omi67 did

The accurate number of years is calculated as follows:

m(t)=85e^(-t/29)

85/10 = 85 * 2 (-t/29), solve for t

t=96.34

Solve for t over the real numbers:
17/2 = 85 2^(-t/29)

17/2 = 85 2^(-t/29) is equivalent to 85 2^(-t/29) = 17/2:
85 2^(-t/29) = 17/2

Divide both sides by 85:
2^(-t/29) = 1/10

Take reciporicals of both sides:
2^(t/29) = 10

Take the logarithm base 2 of both sides:
t/29 = (log(10))/(log(2))

Multiply both sides by 29:
Answer: |t = (29 log(10))/(log(2))=96.34 years.

Omi67:

See my explanation here:

http://web2.0calc.com/questions/m-t-85e-t-409-describes-the-amount-of-a-radioactive-isotope-as-a-function-of-time-t-years-after-how-long-the-amount-has-dropped-to-1-6-of-the

5^(2*x) = 25

5^(2*x)=5^2 Equate the exponents

2*x =2 divide both sides by 2

x =2/2

x=1

I have a different result.

Solve for x over the real numbers:
(5+e^x)/(4+9/e^x) = 2

Simplify and substitute y = e^x:
 (5+e^x)/(4+9/e^x) = (e^x (5+e^x))/(9+4 e^x) = (y (y+5))/(4 y+9) = 2:
(y (y+5))/(4 y+9) = 2

Multiply both sides by 4 y+9:
y (y+5) = 2 (4 y+9)

Expand out terms of the left hand side:
y^2+5 y = 2 (4 y+9)

Expand out terms of the right hand side:
y^2+5 y = 8 y+18

Subtract 8 y+18 from both sides:
y^2-3 y-18 = 0

The left hand side factors into a product with two terms:
(y-6) (y+3) = 0

Split into two equations:
y-6 = 0 or y+3 = 0

Add 6 to both sides:
y = 6 or y+3 = 0

Substitute back for y = e^x:
e^x = 6 or y+3 = 0

Take the natural logarithm of both sides:
x = log(6) or y+3 = 0

Subtract 3 from both sides:
x = log(6) or y = -3

Substitute back for y = e^x:
x = log(6) or e^x = -3

e^x = -3 has no solution since for all z element R, e^z>0 and -3<0:
Answer: |x = log(6)

first off you spelled "peace" wrong it is piece  and the answer i came up with is 4.

M(t)=85e^(-t/719)

85/6 =85 * 2^(-t/719), solve for t

t=1,858.59 years.

Solve for t over the real numbers:
85/6 = 85/2^(t/719)

85/6 = 85/2^(t/719) is equivalent to 85/2^(t/719) = 85/6:
85/2^(t/719) = 85/6

Divide both sides by 85:
2^(-t/719) = 1/6

Take reciporicals of both sides:
2^(t/719) = 6

Take the logarithm base 2 of both sides:
t/719 = (log(6))/(log(2))

Multiply both sides by 719:
Answer: |t = (719 log(6))/(log(2))=1,858.59 years.

CPhill: When you know the half-life of a radioactive material, first you have to figure out its "k"=constant of decay=ln(1/2) / 409 =-0.0016947363......Then the equation becomes:

85/6 =85* e^(-0.0016947363t), solve for t:

t=1,057.25 years.

If you know the half-life, as is the case here, then this procedure is used, which gives the same answer:85/6 =85 * 2^(-t/409), solve for t:

t=1,057.25 years.

It makes sense to notice that since the half-life is 409 years, it will take about 2.58 x 409=1,057.25 to go down to 1/6 of the original amount. At 732.83 years, it will go down to 2^(-732.82/409)=29% of the original amout or about 24.55 units.

what do you mean?

Show me further down.