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Drag each label to the correct location in the equation. Each label can be used more than once, but not all labels will be used.

The number of bacteria, [tex]\( b \)[/tex], in a refrigerated food is given by the function [tex]\( b(t) = 15t^2 - 60t + 125 \)[/tex], where [tex]\( t \)[/tex] is the temperature of the food in degrees Fahrenheit. The function [tex]\( t(h) = 3h + 4 \)[/tex] gives the temperature, [tex]\( t \)[/tex], of the food [tex]\( h \)[/tex] hours after being removed from the refrigerator. Find the number of bacteria in the food in [tex]\( h \)[/tex] hours.

[tex]\[
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline
$b \circ t(h)$ & $=$ & & $h^2$ & & & $h$ & \\
\hline
\end{tabular}
\][/tex]

Labels:
135, 180, 125, 160, 155



Answer :

GinnyAnswer
Let's break down the calculation step by step and find the correct labeling for the given equation, finding the number of bacteria in the food in [tex]$h$[/tex] hours.

We have two functions:
1. The number of bacteria in terms of temperature [tex]\( t \)[/tex]:
[tex]\[ b(t) = 15t^2 - 60t + 125 \][/tex]
2. The temperature in terms of hours [tex]\( h \)[/tex]:
[tex]\[ t(h) = 3h + 4 \][/tex]

Firstly, we are going to substitute [tex]\( t(h) \)[/tex] into [tex]\( b(t) \)[/tex] to find [tex]\( b(t(h)) \)[/tex].

Given:
[tex]\[ t(h) = 3h + 4 \][/tex]

We substitute [tex]\( t(h) \)[/tex] into [tex]\( b(t) \)[/tex]:

[tex]\[ b(t(h)) = b(3h + 4) \][/tex]

Now, we substitute [tex]\( 3h + 4 \)[/tex] into the equation for bacteria:
[tex]\[ b(3h + 4) = 15(3h + 4)^2 - 60(3h + 4) + 125 \][/tex]

Next, expand and simplify the expression:
[tex]\[ (3h + 4)^2 = (3h)^2 + 2 \cdot 3h \cdot 4 + 4^2 = 9h^2 + 24h + 16 \][/tex]

Substitute this back into the bacterial function:
[tex]\[ b(3h + 4) = 15(9h^2 + 24h + 16) - 60(3h + 4) + 125 \][/tex]

Distribute the 15 and simplify:
[tex]\[ = 135h^2 + 360h + 240 - 180h - 240 + 125 \][/tex]

Combine like terms:
[tex]\[ = 135h^2 + 180h + 125 \][/tex]

Therefore, the function for the number of bacteria in [tex]\( h \)[/tex] hours is:
[tex]\[ b(t(h)) = 135h^2 + 180h + 125 \][/tex]

Using the labels provided, we can fill in the equation accordingly:
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline
[tex]$b \circ t(h)$[/tex] & [tex]$=$[/tex] & 135 & [tex]$h^2$[/tex] & + & 180 & [tex]$h$[/tex] & + 125 \\
\hline
\end{tabular}

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