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mojhsa [17]
3 years ago
11

The function f(x) = 15(2)^x represents the growth of a frog population every year in a remote swamp. Elizabeth wants to manipula

te the formula to an equivalent form that calculates every half-year, not every year. Which function is correct for Elizabeth's purposes?
f(x) = 15(2^1/2)^2x
f(x) = 15(2^2) ^x/2
f(x) = 15/2(2)^x
f(x) = 30(2)^x
Mathematics
2 answers:
weeeeeb [17]3 years ago
5 0
Half one is 1/2;
half three is 3/2;
Thus, if f(x) represents annual growth, f(x)/2 shows it every half-year, like this:
\dfrac{f(x)}{2} = \dfrac{15\cdot 2^x}{2} \\  \\ 
\dfrac{f(x)}{2} = \dfrac{15}{2}\cdot 2^x

Written in plain text: f(x)=15/2*2^x
vaieri [72.5K]3 years ago
5 0

Answer:

\boxed{\boxed{B.\ f(x)=15(2^2)^{\frac{x}{2}}}}

Step-by-step explanation:

The general equation for exponential growth is,

y=a(1+r)^x

where,

a = initial amount,

r = rate of growth,

y = future amount,

x = time.

The function f(x) = 15(2)^x represents the growth of a frog population every year in a remote swamp.

where 15 is the initial amount of frog.

As Elizabeth wants to manipulate the formula to an equivalent form that calculates every half-year, so putting x=\dfrac{x}{2} in the general function,

y=15(1+r)^{\frac{x}{2}}

Now we have to find the value of r.

As we will get same future amount, irrespective to the function used, so comparing the old function with the new function,

\Rightarrow 15(1+r)^{\frac{x}{2}}=15(2)^x

Multiplying the exponents of both sides by \dfrac{1}{x}

\Rightarrow (1+r)^{\frac{x}{2}\times \frac{1}{x}}=(2)^{x\times \frac{1}{x}}

\Rightarrow (1+r)^{\frac{1}{2}}=(2)^1

Squaring both sides,

\Rightarrow (1+r)=(2)^2=4

\Rightarrow r=4-1=3

Putting the value of r, in the general equation,

y=15(1+3)^{\frac{x}{2}}=15(4)^{\frac{x}{2}}=15(2^2)^{\frac{x}{2}}


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