Answer:
the remaining mass of unobtanium -41 after a 5 seconds, 10 seconds, 15 seconds and 20 seconds is 24 gr, 12 gr , 5.33 gr and 3 gr respectively
Step-by-step explanation:
the equation that governs the remaining mass m of unobtanium -41 after a time t is
m=m₀*2^(-t/T) , where t is in seconds ,m₀ represents initial mass and T=half-life
Therefore
a) for t=5 s
m=48 gr*2^(-t/5 s) = 48 gr*2^(-5 s/5 s) = 48 gr/2 = 24 gr
b) for t=10 s
m=48 gr*2^(-t/5 s) = 48 gr*2^(-10 s/5 s) = 48 gr/4 = 12 gr
b) for t=15 s
m=48 gr*2^(-t/5 s) = 48 gr*2^(-15 s/5 s) = 48 gr/9 = 5.33 gr
b) for t=20 s
m=48 gr*2^(-t/5 s) = 48 gr*2^(-20 s/5 s) = 48 gr/16 = 3 gr
thus the remaining mass of unobtanium -41 after a 5 seconds, 10 seconds, 15 seconds and 20 seconds is 24 gr, 12 gr , 5.33 gr and 3 gr respectively
Answer:
-8x + 146
Step-by-step explanation:
Step 1: Write out expression
3(58 - 6) - 2(4x + 5)
Step 2: Parenthesis
3(52) - 2(4x + 5)
Step 3: Multiply
156 - 2(4x + 5)
Step 4: Distribute
156 - 8x - 10
Step 5: Combine like terms (constants)
-8x + 146
