Which equation represents an exponential function that passes through the point (2, 80)?
1 answer:
The function f(x) = 5(x)⁴, and f(x) = 5(4)ˣ represents the function that passes through the point (2, 80) option second and fourth are correct.
<h3>What is an exponential function?</h3>It is defined as the function that rapidly increases and the value of the exponential function is always a positive. It denotes with exponent
where a is a constant and a>1
We have; an exponential function that passes through the point (2, 80).
f(x) = 4(x)⁵
Plug (2, 80) in the abovew function:
f(2) = 4(2)⁵ = 128 (false)
f(x) = 5(x)⁴
Plug (2, 80) in the abovew function:
f(2) = 5(2)⁴ = 80 (true)
f(x) = 4(5)ˣ
f(2) = 4(5)² = 100 (false)
f(x) = 5(4)ˣ
f(2) = 5(4)² = 80 (true)
Thus, the function f(x) = 5(x)⁴, and f(x) = 5(4)ˣ represents the function that passes through the point (2, 80) option second and fourth are correct.
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Answer:
Step-by-step explanation:
We want to find the inverse of the function:
To find the inverse of a function, we will flip the variables and solve for <em>y</em>. So:
Solve for <em>y</em>. Subtract two from both sides:
Divide both sides by two:
And take the principal square root of both sides. Therefore:
Answer:
6 billion years.
Step-by-step explanation:
According to the decay law, the amount of the radioactive substance that decays is proportional to each instant to the amount of substance present. Let be the amount of
and
be the amount of
after
years.
Then, we obtain two differential equations
where and
are proportionality constants and the minus signs denotes decay.
Rearranging terms in the equations gives
Now, the variables are separated, and
appear only on the left, and
appears only on the right, so that we can integrate both sides.
which yields
,
where and
are constants of integration.
By taking exponents, we obtain
Hence,
,
where and
.
Since the amounts of the uranium isotopes were the same initially, we obtain the initial condition
Substituting 0 for in the general solution gives
Similarly, we obtain and
The relation between the decay constant and the half-life is given by
We can use this fact to determine the numeric values of the decay constants and
. Thus,
and
Therefore,
We have that
Hence,
Solving for yields
, which means that the age of the universe is about 6 billion years.