5. let ( a_n ) be a geometric sequence with ( a_2 = 42 ) and ( a_5 = 141.75 ). find ( \frac{a_3}{a_2} ).
6. values of a function ( f ) for selected values of ( x ) are given in the table.
\begin{array}{c|c|c|c}x & 3 & 6 & 9 \hlinef(x) & k & 10 & 80 \end{array}
a. find the value of ( k ) if ( f ) is a linear function.
b. find the value of ( k ) if ( f ) is an exponential function.
7. calliope is taking a 14 - day road trip out west. on the first day of her trip, she travels 40 miles. on the second day, she travels 55 miles. on the third day, she travels 70 miles and on the fourth day she travels 85 miles. she continues in this pattern for all 14 days.
a. how many miles does she drive on day 7?
b. write a rule for the number of miles she travels on the ( n^{ ext{th}} ) day of her road trip.
c. how many miles does she drive on day 14?
d. how many total miles did she drive during the second week of her trip (days 8 – 14)? explain your reasoning.

Question

5. let ( a_n ) be a geometric sequence with ( a_2 = 42 ) and ( a_5 = 141.75 ). find ( \frac{a_3}{a_2} ).
6. values of a function ( f ) for selected values of ( x ) are given in the table.
\begin{array}{c|c|c|c}x & 3 & 6 & 9 \hlinef(x) & k & 10 & 80 \end{array}
a. find the value of ( k ) if ( f ) is a linear function.
b. find the value of ( k ) if ( f ) is an exponential function.
7. calliope is taking a 14 - day road trip out west. on the first day of her trip, she travels 40 miles. on the second day, she travels 55 miles. on the third day, she travels 70 miles and on the fourth day she travels 85 miles. she continues in this pattern for all 14 days.
a. how many miles does she drive on day 7?
b. write a rule for the number of miles she travels on the ( n^{	ext{th}} ) day of her road trip.
c. how many miles does she drive on day 14?
d. how many total miles did she drive during the second week of her trip (days 8 – 14)? explain your reasoning.

Accepted Answer

### Problem 5
# Explanation:
## Step1: Recall geometric sequence formula
In a geometric sequence, $ a_n = a_1 r^{n - 1} $, where $ a_1 $ is the first term and $ r $ is the common ratio. So $ a_2 = a_1 r^{2 - 1}=a_1 r = 42 $, and $ a_5 = a_1 r^{5 - 1}=a_1 r^4 = 141.75 $.

## Step2: Find the common ratio $ r $
Divide $ a_5 $ by $ a_2 $: $ \frac{a_5}{a_2}=\frac{a_1 r^4}{a_1 r}=\frac{r^4}{r}=r^3 $. Substitute the values: $ \frac{141.75}{42}=r^3 $. Calculate $ \frac{141.75}{42}=3.375 $. So $ r^3 = 3.375 $, and $ r=\sqrt[3]{3.375}=1.5 $ (since $ 1.5^3 = 3.375 $).

## Step3: Find $ \frac{a_3}{a_2} $
In a geometric sequence, $ \frac{a_{n + 1}}{a_n}=r $. So $ \frac{a_3}{a_2}=r $. We found $ r = 1.5=\frac{3}{2} $.

# Answer:
$ \frac{3}{2} $ (or $ 1.5 $)

### Problem 6a
# Explanation:
## Step1: Recall linear function formula
A linear function has the form $ f(x)=mx + b $, where $ m $ is the slope. The slope between two points $ (x_1,y_1) $ and $ (x_2,y_2) $ is $ m=\frac{y_2 - y_1}{x_2 - x_1} $.

## Step2: Calculate the slope
Using $ (6,10) $ and $ (9,80) $, $ m=\frac{80 - 10}{9 - 6}=\frac{70}{3} $? Wait, no, wait. Wait, for linear function, the slope between $ (3,k) $, $ (6,10) $ and $ (9,80) $ should be equal. So slope between $ (3,k) $ and $ (6,10) $ is $ \frac{10 - k}{6 - 3}=\frac{10 - k}{3} $, and slope between $ (6,10) $ and $ (9,80) $ is $ \frac{80 - 10}{9 - 6}=\frac{70}{3} $. Set them equal: $ \frac{10 - k}{3}=\frac{70}{3} $. Multiply both sides by 3: $ 10 - k = 70 $. Then $ -k = 70 - 10 = 60 $, so $ k=-60 $. Wait, no, that can't be. Wait, no, wait, maybe I made a mistake. Wait, linear function: the difference in $ y $ over difference in $ x $ is constant. So from $ x = 3 $ to $ x = 6 $, $ \Delta x=3 $, $ \Delta y = 10 - k $. From $ x = 6 $ to $ x = 9 $, $ \Delta x=3 $, $ \Delta y = 80 - 10 = 70 $. So for linear, $ 10 - k = 70 $? No, that would mean $ k = 10 - 70=-60 $. Wait, but let's check. If $ k=-60 $, then from $ x=3 $ (y=-60) to $ x=6 $ (y=10), the slope is $ \frac{10 - (-60)}{6 - 3}=\frac{70}{3}\approx23.33 $. From $ x=6 $ (y=10) to $ x=9 $ (y=80), slope is $ \frac{80 - 10}{9 - 6}=\frac{70}{3}\approx23.33 $. So that works.

# Answer:
$ k=-60 $

### Problem 6b
# Explanation:
## Step1: Recall exponential function formula
An exponential function has the form $ f(x)=a b^x $. So we have $ f(3)=a b^3 = k $, $ f(6)=a b^6 = 10 $, $ f(9)=a b^9 = 80 $.

## Step2: Find $ b $
Divide $ f(9) $ by $ f(6) $: $ \frac{a b^9}{a b^6}=b^3=\frac{80}{10}=8 $. So $ b^3 = 8 $, so $ b = 2 $.

## Step3: Find $ a $
Use $ f(6)=a b^6 = 10 $. Since $ b = 2 $, $ b^6 = 64 $. So $ a\times64 = 10 $, so $ a=\frac{10}{64}=\frac{5}{32} $.

## Step4: Find $ k = f(3) $
$ f(3)=a b^3 $. We know $ a=\frac{5}{32} $, $ b^3 = 8 $. So $ f(3)=\frac{5}{32}\times8=\frac{5}{4}=1.25 $. Wait, no, wait: $ b^3 = 8 $, so $ f(3)=a b^3=\frac{10}{64}\times8=\frac{10}{8}=1.25 $? Wait, no, $ a b^6 = 10 $, $ b^6=(b^3)^2 = 64 $, so $ a=\frac{10}{64}=\frac{5}{32} $. Then $ f(3)=a b^3=\frac{5}{32}\times8=\frac{5}{4}=1.25 $. Wait, but let's check with $ f(6) $: $ a b^6=\frac{5}{32}\times64 = 10 $, which is correct. And $ f(9)=a b^9=\frac{5}{32}\times512 = 80 $, which is correct. So $ k = 1.25=\frac{5}{4} $.

# Answer:
$ k=\frac{5}{4} $ (or $ 1.25 $)

### Problem 7a
# Explanation:
## Step1: Identify the sequence type
The miles per day: 40, 55, 70, 85,... This is an arithmetic sequence with first term $ a_1 = 40 $ and common difference $ d = 15 $ (since $ 55 - 40 = 15 $, $ 70 - 55 = 15 $, etc.).

## Step2: Use arithmetic sequence formula
The $ n $-th term of an arithmetic sequence is $ a_n = a_1+(n - 1)d $. For day 7, $ n = 7 $. So $ a_7 = 40+(7 - 1)\times15 = 40 + 6\times15 = 40 + 90 = 130 $.

# Answer:
130

### Problem 7b
# Explanation:
## Step1: Recall arithmetic sequence formula
For an arithmetic sequence, the $ n $-th term is $ a_n = a_1+(n - 1)d $. We have $ a_1 = 40 $, $ d = 15 $.

## Step2: Write the rule
Substitute $ a_1 $ and $ d $ into the formula: $ a_n = 40+(n - 1)\times15 $. Simplify: $ a_n = 40 + 15n - 15 = 15n + 25 $.

# Answer:
$ a_n = 15n + 25 $ (or $ a_n = 40 + 15(n - 1) $)

### Problem 7c
# Explanation:
## Step1: Use the arithmetic sequence formula
We have $ a_n = 15n + 25 $ (from part b). For day 14, $ n = 14 $. So $ a_{14}=15\times14 + 25 = 210 + 25 = 235 $.

# Answer:
235

### Problem 7d
# Explanation:
## Step1: Identify the terms for days 8 - 14
These are 7 terms (days 8,9,10,11,12,13,14). The first term of this sub - sequence (day 8) is $ a_8 $, and the last term is $ a_{14}=235 $ (from part c). First, find $ a_8 $: using $ a_n = 15n + 25 $, $ a_8 = 15\times8 + 25 = 120 + 25 = 145 $.

## Step2: Use the sum formula for arithmetic series
The sum of an arithmetic series is $ S_n=\frac{n(a_1 + a_n)}{2} $, where $ n $ is the number of terms, $ a_1 $ is the first term of the series, and $ a_n $ is the last term. Here, $ n = 7 $, $ a_1 = 145 $, $ a_n = 235 $.

## Step3: Calculate the sum
$ S_7=\frac{7\times(145 + 235)}{2}=\frac{7\times380}{2}=7\times190 = 1330 $.

# Answer:
1330

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 5

Step1: Recall geometric sequence formula

Step2: Find the common ratio \( r \)

Step3: Find \( \frac{a_3}{a_2} \)

Step1: Recall linear function formula

Step2: Calculate the slope

Step1: Recall exponential function formula

Step2: Find \( b \)

Step3: Find \( a \)

Step4: Find \( k = f(3) \)

Answer:

Problem 6a