The number of bacteria triples every 5 days. If the bacteria population starts with 10 bacteria, then the equation

Question

A(t) = 10(3)^0.2t can represent the number of bacteria. Find the value for A(15).

Answer 1

To find the value of \( A(15) \) using the equation \( A(t) = 10(3)^{0.2t} \), we first substitute \( t = 15 \) into the equation:

\[ A(15) = 10(3)^{0.2 \times 15} \]

Calculating the exponent:

\[ 0.2 \times 15 = 3 \]

Now substitute back into the equation:

\[ A(15) = 10(3)^3 \]

Next, we calculate \( (3)^3 \):

\[ 3^3 = 27 \]

Now, substitute this value back into the equation:

\[ A(15) = 10 \times 27 = 270 \]

Thus, the value for \( A(15) \) is

\[ \boxed{270} \]

Answer 2

Q2 Complete the table given the function f(x)=-2(x-1)2+5

x
f(x)
-2

-1

0

1

2

Answer 3

To complete the table for the function \( f(x) = -2(x - 1)^2 + 5 \), we will calculate \( f(x) \) for each given value of \( x \):

1. For \( x = -2 \): \[ f(-2) = -2(-2 - 1)^2 + 5 = -2(-3)^2 + 5 = -2(9) + 5 = -18 + 5 = -13 \]

2. For \( x = -1 \): \[ f(-1) = -2(-1 - 1)^2 + 5 = -2(-2)^2 + 5 = -2(4) + 5 = -8 + 5 = -3 \]

3. For \( x = 0 \): \[ f(0) = -2(0 - 1)^2 + 5 = -2(-1)^2 + 5 = -2(1) + 5 = -2 + 5 = 3 \]

4. For \( x = 1 \): \[ f(1) = -2(1 - 1)^2 + 5 = -2(0)^2 + 5 = -2(0) + 5 = 0 + 5 = 5 \]

5. For \( x = 2 \): \[ f(2) = -2(2 - 1)^2 + 5 = -2(1)^2 + 5 = -2(1) + 5 = -2 + 5 = 3 \]

Now, we can summarize the results in a table:

\[ \begin{array}{|c|c|} \hline x & f(x) \ \hline -2 & -13 \ -1 & -3 \ 0 & 3 \ 1 & 5 \ 2 & 3 \ \hline \end{array} \]

So the completed table is:

\[ \begin{array}{|c|c|} \hline x & f(x) \ \hline -2 & -13 \ -1 & -3 \ 0 & 3 \ 1 & 5 \ 2 & 3 \ \hline \end{array} \]