The half-life of cesium-137 is 30 years. Suppose we have a 20 mg sample. a. Find the mass (in mg) that remains after t years.
b. How much of the sample (in mg) remains after 20 years?
c. After how long will only 1 mg remain?

(a) A = (20mg)/(2^(t/30))

(b) 12.6mg

(c) 129.6years

Step-by-step explanation:

To calculate the amount remaining after a number of half-lives, n, we can make use of:

Where A = amount remaining

B = initial amount

(a) A = (20mg)/(2^(t/30))

(b) Mass after 20years

A = (20mg)/(2^(20/30)) ≈ 12.6mg

(c) After how long will only 1mg remain:

1mg = (20mg)/(2^(t/30))

Taking log of both sides we have:

Log(20) = (t/30)log(2)

t/30 = (log(20))/(log(2)) ≈ 4.3

t/30 = 4.3

t = 30 x 4.3 ≈ 129.6years.

there are many solutions.

step-by-step explanation:

step 1: simplify both sides of the equation.

4x+6=4x^2+10x−4

step 2: subtract 4x^2+10x-4 from both sides.

4x+6−(4x^2+10x−4)=4x^2+10x−4−(4x^2+10x−4)

−4x^2−6x+10=0

step 3: factor left side of equation.

−2(2x+5)(x−1)=0

step 4: set factors equal to 0.

x=-5/2 or x=1

i have no idea just need points sorry

step-by-step explanation: