Answer:
Given that,
The number of grams A of a certain radioactive substance present at time, in years
from the present, t is given by the formula

a) To find the initial amount of this substance
At t=0, we get


We know that e^0=1 ( anything to the power zero is 1)
we get,

The initial amount of the substance is 45 grams
b)To find thehalf-life of this substance
To find t when the substance becames half the amount.
A=45/2
Substitute this we get,


Taking natural logarithm on both sides we get,







Half-life of this substance is 154.02
c) To find the amount of substance will be present around in 2500 years
Put t=2500
we get,




The amount of substance will be present around in 2500 years is 0.000585 grams
Answer:
y=15
Step-by-step explanation:
Answer:
24.56
Step-by-step explanation:
1.14x-5+5=23+5
1.14x=28
x is roughly equal to 24.56.
Hope this helps!
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Information Given:
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ON = 7x - 9
LM = 6x + 4
MN = x - 7
OL = 2y - 7
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Since it is a parallelogram:
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ON = LM and
MN = OL
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ON = LM:
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7x - 9 = 6x + 4
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Subtract 6x from both sides:
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x - 9 = 4
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Add 9 to both sides:
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x = 13
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MN = OL:
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x - 7 = 2y - 7
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Sub x = 13:
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13 - 7 = 2y - 7
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Simplify:
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6 = 2y - 7
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Add 7 on both sides:
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13 = 2y
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Divide by 2:
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y = 13/2
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Answer: x = 13, y = 13/2 (Answer D)
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