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adelina 88 [10]
3 years ago
11

Help with math problem

Mathematics
1 answer:
shusha [124]3 years ago
3 0
What is your math problem?
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A certain substance doubles its volume every minute at 9 AM a small amount is placed in a container ,at 10 AM the container was
Y_Kistochka [10]

Answer:

the container is 1/4 full at 9:58 AM

Step-by-step explanation:

since the volume doubles every minute , the formula for calculating the volume V at any time t is

V(t)=V₀*2^-t , where t is in minutes back from 10 AM and V₀= container volume

thus for t=1 min (9:59 AM) the volume is V₁=V₀/2 (half of the initial one) , for t=2 (9:58 AM) is V₂=V₁/2=V₀/4  ...

therefore when the container is 1/4 full the volume is V=V₀/4 , thus replacing in the equation we obtain

V=V₀*2^-t

V₀/4 = V₀*2^-t

1/4 = 2^-t

appling logarithms

ln (1/4) = -t* ln 2

t = - ln (1/4)/ln 2 = ln 4 /ln 2 = 2*ln 2 / ln 2 = 2

thus t=2 min before 10 AM → 9:58 AM

therefore the container is 1/4 full at 9:58 AM

8 0
3 years ago
Please help!!...............................
Airida [17]
B. you have to be ready to multiply all the numbers to get your answer.
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What is the remainder of (x 3 - 6x 2 -9x + 3) ÷ (x - 3)?
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