The function e(t)=4^8*4^t approximates the number of nematodes in a certain sample of fresh compost after t days. find the initial number of nematodes when t=0. how many nematodes are there after 3.5 days? a. 0; 16,777,216 b. 4^8; 8,388,608 c. 4^8; 16,777,216 d. 0; 8,388,608

We are given function E(t) = 4^8*4^t to approximates the number of nematodes in a certain sample of fresh compost,

t is the time in number of days.

We need to find the initial number of nematodes.

We need to take t=0 for find the initial number of nematodes, because initial number of minutes will be 0.

So, we just need to plug t=0 in the given function.

a) Plugging t=0 in given function

E(t) = 4^8*4^t, we get

E(0) = 4^8 * 4^0.

Please note: Power 0 of any number or term always have a value 1.

Therefore, 4^0 = 1.

Plugging this value in above equation.

E(0) = 4^8 * 4^0 = 4^8 *1

E(0) = 4^8.

Therefore, 4^8 number of nematodes are there initially.

b) In second part, we need to find the number of nematodes after 3.5 days.

Plugging t=3.5 in given function, we get

E(3.5) = 4^8 * 4^(3.5).

4^(3.5) = 128 and 4^8= 65536

Therefore,

E(3.5) = 4^8 * 4^(3.5) = 65536 * 128 = 8388608.

Therefore, 8,388,608 nematodes are there after 3.5 days.

And Option B. 4^8; 8,388,608 is correct option.

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