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
If it took you (one person) 3 days to build a doghouse, then it should take 2 people to build it in half the time. So all we have to do is split 3 days in 2.
3/2 = 1.5
So it would take 1 1/2 days or 36 hours for 2 people to complete the doghouse.
The sequence is " Algebraic, common difference = −10 " ⇒ 1st answer
Step-by-step explanation:
Let us revise the algebraic and geometric sequences
- The Algebraic sequence is the sequence that has a common difference between each two consecutive terms, like 2 , 5 , 8 , 11 , 14 , .......... the difference between the 5 and 2 is equal to the difference between 8 and 5, and the difference between 11 and 8 and the difference between 14 and 11
- The geometric sequences is the sequence that has a common ratio between each two consecutive terms, like 3 , -9 , 27 , -81 , ...... the ratio between -9 and 3 as the ratio between 27 and -9 as the ratio between -81 and 27
→ x : 1 : 2 : 3 : 4
→ f(x) : 5 : -5 : -15 : -25
x represents the positions of the terms
f(x) represents the value of each term
∵ 1st = 5 and 2nd = -5
∵ -5 - 5 = -10
∵ 2nd = -5 and 3rd = -15
∵ -15 - (-5) = -15 + 5 = -10
∵ 3rd = -15 and 4th = -25
∵ -25 - (-15) = -25 + 15 = -10
∴ There is a common difference -10 between each two
consecutive terms
∴ The sequence is algebraic with common difference -10
The sequence is " Algebraic, common difference = −10 "
Learn more:
You can learn more about the sequences in brainly.com/question/1522572
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