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Anything except zero to the zero power is 1.

thank you so much 

Hey :D

Who's TannerFox?

NO!. When you calculate k, you no longer use the half-life given, which is 719 years in this case. So that your equation should read: 85/6 =85 * e^(-0.000964 * t), which is equal to: 85/6 =85 * 2^(-t/719). If you now solve for t in both cases, you will get EXACTLY the same answer, which comes to:1,858.59 years.

That's not an inequality.

\(8/3 - 6 = 8/3 - 18/3 = (8-18)/3 = -10/3\)

\(\ln (5^{2x})=\ln 25\\ 2x \ln 5 = \ln 25\\ x \ln 5^2 = \ln 25\\ x = \ln25 / \ln 25 = 1\)

\(\text{Example:} \sqrt[3]{8}\\ \text{You identify }8 = 2^3 \sqrt[3]{8}\\ = \sqrt[3]{2^3}\\ = 2\)

Remembering that 1/4 = 0.25.....

5/4 = 1/4 x 5 = 0.25 x 5 = 1.25

The angle formed by the diagonal and the side will always be 45 degrees.

sin 45 = 7 / diagonal

diagonal = 7sqrt(2)

a(0)=50

a(1)=1/2 x 50 + 10 = 35

a(2)= 1/2 x 35 + 10 = 55/2

a(3)= 1/2 x  27.5 + 10 =75/4

a(n) = \(50 - 30\overbrace{(1/2+1/4+1/8+.......)}^{n}\)= \(50 - 30(\dfrac{2^n-1}{2^n})\)

1/6 (85) = 85 e^(-t/719)         this is of the form   A = Ao e^kt    where k= -.0013908

1/6 = e^(-t/719)

ln(1/6) = (-t/719)

-719 ln(1/6) = t     = 1288.28 years

NO!!. 29 IS HALF-LIFE!! Half-life of any radioactive substance is calculated as: ln(1/2) / k, when k is known. IF it is not known, then it is calculated as: ln(1/2) / half-life of the substance.

I think you should have Alan explain it to you if you don't understand the technical page I referred you to. http://math.usask.ca/emr/examples/expdeceg.html

P.S. The whole problem stems from the way the question is posed. When e is used, it automatically implies that you have to calculate k, so that it should read: e^(kt). When they give you the half-life, it automatically implies that you must calculate k. You can get the same result by using: 2^(-t/half-life), or: 1/2 ^ (t/half-life).

No.

Use pythagorean theorem   a^2 + b^2 = c^2      where c is the diagonal   and  a and b are the side lengths (both 7)

7^2 + 7^2 = c^2

49 + 49 = 98 = c^2    c = sqrt(98) = 9.899 in

1.25

(8+8)- 8^0 - 8^0

can someone post part b? im having trouble

K = -1/29    IS less than  0 

Ao = Ao e^(-t/29)

1/2Ao = e^(-t/29)

ln 1/2 = -t/29

ln 1/2  *  29  =  -t

t=20.10     is the half life

85   one half life  42.5    two half lives  21.25   three half lives  10.6.5      down to 8.5 is another partial half life

            20.1                 +     20.1                     +      20.1                          +     ~6      =  ~   66.3  years