Math / Algebra

QuestionLesson 6 Checkpoint EVALUATE
Independent Practice \begin{tabular}{|l|c|c|} \hline Learning Goals & Lesson Reflection (circle one) \\ \hline \begin{tabular}{l} I can determine an explicit expression or steps for \\ calculation from a context \end{tabular} & Starting... Getting There... Got it! \\ \hline \end{tabular} Complete the previous problems, check your solutions, then complete the Lesson Checkpoint below. Complete the Lesson Reflection above by circling your current understanding of the Learning Goal(s).
Write an explicit rule in function notation for the arithmetic sequence.
1. A student loan needs to be paid off beginning the first year after graduation. Beginning at Year 1 , there is $\$ 52,000$ remaining to be paid. The graduate makes regular payments of $\$ 8,000$ each year. The graph shows the sequence.
A home cook needs to boil 1 liter of water on the stove. Water boils at $100^{\circ} \mathrm{C}$ . \begin{tabular}{|l|c|c|c|} \hline Minutes & 3 & 4 & 5 \\ \hline Temperature $\left({ }^{\circ} \mathrm{C}\right)$ & 38 & 46 & 54 \\ \hline \end{tabular}
2. What is the common difference of the sequence represented in the table?
3. What is the explicit rule for the sequence?
4. Find the temperature at 10 minutes. Is the water boiling yet? lifelong Algebra 1A (2024) Module 3

Studdy Solution

STEP 1

What is this asking? We're figuring out a formula to predict the water temperature at any given time, and then we'll use it to see if the water is boiling after 10 minutes. Watch out! Make sure to check the units – we're dealing with minutes and degrees Celsius.
Also, don't forget that water boils at $100^{\circ} \mathrm{C}$ .

STEP 2

1. Find the common difference.
2. Create the explicit rule.
3. Calculate the temperature at 10 minutes.

STEP 3

Let's look at the change in temperature between the given times.
From 3 minutes to 4 minutes, the temperature increases by $46^{\circ} \mathrm{C} - 38^{\circ} \mathrm{C} = 8^{\circ} \mathrm{C}$ .

STEP 4

From 4 minutes to 5 minutes, the temperature increases by $54^{\circ} \mathrm{C} - 46^{\circ} \mathrm{C} = 8^{\circ} \mathrm{C}$ .
So, the common difference is $8^{\circ} \mathrm{C}$ per minute!

STEP 5

We know the common difference is $8^{\circ} \mathrm{C}$ .
We can represent the temperature at any given minute $n$ with a formula like this: $f(n) = \text{starting temperature} + \text{common difference} \cdot n$ .
But what's the starting temperature?

STEP 6

We don't know the temperature at 0 minutes, but we do know the temperature at 3 minutes is $38^{\circ} \mathrm{C}$ .
If the temperature increases by $8^{\circ} \mathrm{C}$ each minute, then 3 minutes before the 3-minute mark (which is 0 minutes), the temperature would have been $38^{\circ} \mathrm{C} - 3 \cdot 8^{\circ} \mathrm{C} = 38^{\circ} \mathrm{C} - 24^{\circ} \mathrm{C} = 14^{\circ} \mathrm{C}$ .
So our initial temperature, $f(0)$ , is $14^{\circ} \mathrm{C}$ !

STEP 7

Now we can write our explicit rule: $f(n) = 14 + 8n$ , where $n$ is the number of minutes and $f(n)$ is the temperature in degrees Celsius.

STEP 8

Using our awesome formula, we can plug in $n = 10$ to find the temperature at 10 minutes: $f(10) = 14 + 8 \cdot 10 = 14 + 80 = 94$ .

STEP 9

So, after 10 minutes, the water temperature is $94^{\circ} \mathrm{C}$ .
Since water boils at $100^{\circ} \mathrm{C}$ , the water is not boiling yet!

STEP 10

The explicit rule for the sequence is $f(n) = 14 + 8n$ .
The temperature at 10 minutes is $94^{\circ} \mathrm{C}$ , which means the water is not boiling yet.

Was this helpful?

Studdy Solution

STEP 1

STEP 2

1. Find the common difference.
2. Create the explicit rule.
3. Calculate the temperature at 10 minutes.

STEP 3

Let's look at the change in temperature between the given times.
From 3 minutes to 4 minutes, the temperature increases by $46^{\circ} \mathrm{C} - 38^{\circ} \mathrm{C} = 8^{\circ} \mathrm{C}$ .

STEP 4

From 4 minutes to 5 minutes, the temperature increases by $54^{\circ} \mathrm{C} - 46^{\circ} \mathrm{C} = 8^{\circ} \mathrm{C}$ .
So, the common difference is $8^{\circ} \mathrm{C}$ per minute!

STEP 5

We know the common difference is $8^{\circ} \mathrm{C}$ .
We can represent the temperature at any given minute $n$ with a formula like this: $f(n) = \text{starting temperature} + \text{common difference} \cdot n$ .
But what's the starting temperature?

STEP 6

STEP 7

Now we can write our explicit rule: $f(n) = 14 + 8n$ , where $n$ is the number of minutes and $f(n)$ is the temperature in degrees Celsius.

STEP 8

Using our awesome formula, we can plug in $n = 10$ to find the temperature at 10 minutes: $f(10) = 14 + 8 \cdot 10 = 14 + 80 = 94$ .

STEP 9

So, after 10 minutes, the water temperature is $94^{\circ} \mathrm{C}$ .
Since water boils at $100^{\circ} \mathrm{C}$ , the water is not boiling yet!

STEP 10

The explicit rule for the sequence is $f(n) = 14 + 8n$ .
The temperature at 10 minutes is $94^{\circ} \mathrm{C}$ , which means the water is not boiling yet.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

STEP 2

1. Find the common difference.2. Create the explicit rule.3. Calculate the temperature at 10 minutes.

STEP 3

Let's look at the change in temperature between the given times.From 3 minutes to 4 minutes, the temperature increases by 46∘C−38∘C=8∘C46^{\circ} \mathrm{C} - 38^{\circ} \mathrm{C} = 8^{\circ} \mathrm{C}46∘C−38∘C=8∘C.

STEP 4

From 4 minutes to 5 minutes, the temperature increases by 54∘C−46∘C=8∘C54^{\circ} \mathrm{C} - 46^{\circ} \mathrm{C} = 8^{\circ} \mathrm{C}54∘C−46∘C=8∘C.So, the **common difference** is 8∘C8^{\circ} \mathrm{C}8∘C per minute!

STEP 5

STEP 6

STEP 7

Now we can write our **explicit rule**: f(n)=14+8nf(n) = 14 + 8nf(n)=14+8n, where nnn is the number of minutes and f(n)f(n)f(n) is the temperature in degrees Celsius.

STEP 8

Using our awesome formula, we can plug in n=10n = 10n=10 to find the temperature at 10 minutes: f(10)=14+8⋅10=14+80=94f(10) = 14 + 8 \cdot 10 = 14 + 80 = 94f(10)=14+8⋅10=14+80=94.

STEP 9

So, after 10 minutes, the water temperature is 94∘C94^{\circ} \mathrm{C}94∘C.Since water boils at 100∘C100^{\circ} \mathrm{C}100∘C, the water is *not* boiling yet!

STEP 10

The explicit rule for the sequence is f(n)=14+8nf(n) = 14 + 8nf(n)=14+8n.The temperature at 10 minutes is 94∘C94^{\circ} \mathrm{C}94∘C, which means the water is not boiling yet.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Find the common difference.
2. Create the explicit rule.
3. Calculate the temperature at 10 minutes.

Let's look at the change in temperature between the given times.
From 3 minutes to 4 minutes, the temperature increases by $46^{\circ} \mathrm{C} - 38^{\circ} \mathrm{C} = 8^{\circ} \mathrm{C}$ .

From 4 minutes to 5 minutes, the temperature increases by $54^{\circ} \mathrm{C} - 46^{\circ} \mathrm{C} = 8^{\circ} \mathrm{C}$ .
So, the common difference is $8^{\circ} \mathrm{C}$ per minute!

Now we can write our explicit rule: $f(n) = 14 + 8n$ , where $n$ is the number of minutes and $f(n)$ is the temperature in degrees Celsius.

Using our awesome formula, we can plug in $n = 10$ to find the temperature at 10 minutes: $f(10) = 14 + 8 \cdot 10 = 14 + 80 = 94$ .

So, after 10 minutes, the water temperature is $94^{\circ} \mathrm{C}$ .
Since water boils at $100^{\circ} \mathrm{C}$ , the water is not boiling yet!

The explicit rule for the sequence is $f(n) = 14 + 8n$ .
The temperature at 10 minutes is $94^{\circ} \mathrm{C}$ , which means the water is not boiling yet.