PLEASE HELP The half-life of phosphorous-32 is approximately 24.3 days.
How much of a 3 gram sample of phosphorus-32 would remain after 10 days?
ROUND TO THREE DECIMAL POINTS.

Step-by-step explanation:

We will use the standard formula for half-life:  

Where A₀ is the initial amount, a is the half-life, d is the time for one half-life, and t is the number of days.

So, we will substitute 1/2 for a and 24.3 for d. This yields:

We have a 3 gram sample. Therefore, our initial amount A₀ is 3:

And we want to find the remaining sample after 10 days. So, t=10. Substitute:

Use a calculator. So:

Therefore, for a 3 gram sample of Phosphorous-32 with a half-life of 24.3 days, after 10 days, there will be approximately 2.255 grams remaining.

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.