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user100 [1]
3 years ago
6

A radio active isotope decays according to the exponential decay equation where t is in days. Round to the thousandths place. Th

e half life is the solution (t) of the equation: a/2=ae^-2.360t
Mathematics
1 answer:
Whitepunk [10]3 years ago
8 0
To calculate the value of the half life we proceed as follows;
a/2=ae^(-2.360t)
the a(s) will cancel and we shall remain with:
1/2=e^-2.360t
introducing the natural logs we get:
ln (1/2)=ln e^(-2.360t)
ln(1/2)=-2.360t
hence;
t=(ln 0.5)/(-2.360)
t=0.294
the answer is t=0.294
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3 0
3 years ago
The function f(x) is represented by the table below. What are the corresponding values of g(x) for the transformation g(x) = 6f(
g100num [7]

Answer:

Given: The transformation g(x) =6f(x)                  ......[1]

To find the corresponding values of g(x) for the given transformation.

At x = -7, f(x) = 8

then,

By substituting the value of f(x) = 8 in [1] we have

g(x) = 6 \cdot 8 =48

Similarly,

For x= -3 , f(x) = 3, then

By substituting the value of f(x) = 3 in [1] we have

g(x) = 6 \cdot 3 =18

For x= 0, f(x) = -1 , then

By substituting the value of f(x) = -1 in [1] we have

g(x) = 6 \cdot -1 =-6

For x= 2, f(x) = 7, then

By substituting the value of f(x) = 7 in [1] we have

g(x) = 6 \cdot 7=42

For x= 10, f(x) = 5, then

By substituting the value of f(x) = 5 in [1] we have

g(x) = 6 \cdot 5=30

Therefore ; we have the corresponding values of g(x) as shown below;

x                f(x)           g(x)

-7                8                48

-3                3                 18

0                -1               -6

2                 7                42

10                5                30

5 0
3 years ago
Read 2 more answers
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Ivan

the height:

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3 0
3 years ago
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2(x-3)^2+10=82 solve by isolating x
Leviafan [203]
We have the following equation:
 2 (x-3) ^ 2 + 10 = 82
 We must clear x.
 We pass the constant terms to one side of the equation:
 2 (x-3) ^ 2 = 82 - 10
 2 (x-3) ^ 2 = 72
 (x-3) ^ 2 = 72/2
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 Square root to both members:
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x1 = 9
 
x2 = -3
7 0
3 years ago
A log is 20 m long, correct to the nearest metre.
Marat540 [252]

Answer:

222 fence posts

Step-by-step explanation:

Since the log is 20 m long, correct to the nearest metre and has to be cut into fence posts which must be 90 cm long, correct  to the nearest 10 centimetres, there can be x fence posts 90 cm long.

So, we have x × 90 cm = 20 m

x × 0.9 m = 20 m

x = 20 m/0.9 m

x = 222.22

x ≅ 222 fence posts

So, the largest number of fence posts that can possibly be cut from the log is 222 fence posts.

4 0
3 years ago
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