College

Consider the relationship [tex]7r + 8t = 7[/tex].

a. Write the relationship as a function [tex]r = f(t)[/tex].
Enter the exact answer.
[tex]f(t) = \square[/tex]

b. Evaluate [tex]f(-7)[/tex].
[tex]f(-7) = \square[/tex]

c. Solve [tex]f(t) = 49[/tex].
[tex]t = \square[/tex]

Answer :

Sure, let's work through this step-by-step!

### Part a: Write the relationship as a function [tex]\( r = f(t) \)[/tex]

We start with the given equation:
[tex]\[ 7r + 8t = 7 \][/tex]

Our goal is to express [tex]\( r \)[/tex] as a function of [tex]\( t \)[/tex]. To do this, we'll solve for [tex]\( r \)[/tex]:

1. Subtract [tex]\( 8t \)[/tex] from both sides of the equation:
[tex]\[ 7r = 7 - 8t \][/tex]

2. Divide both sides by 7 to isolate [tex]\( r \)[/tex]:
[tex]\[ r = \frac{7 - 8t}{7} \][/tex]

Therefore, the function [tex]\( r \)[/tex] in terms of [tex]\( t \)[/tex] is:
[tex]\[ f(t) = \frac{7 - 8t}{7} \][/tex]

### Part b: Evaluate [tex]\( f(-7) \)[/tex]

To find [tex]\( f(-7) \)[/tex], we'll substitute [tex]\( t = -7 \)[/tex] into the function [tex]\( f(t) \)[/tex]:

[tex]\[ f(-7) = \frac{7 - 8(-7)}{7} \][/tex]

Calculate inside the parentheses first:
[tex]\[ 7 - 8(-7) = 7 + 56 = 63 \][/tex]

Now, divide by 7:
[tex]\[ f(-7) = \frac{63}{7} = 9 \][/tex]

So, [tex]\( f(-7) = 9 \)[/tex].

### Part c: Solve [tex]\( f(t) = 49 \)[/tex]

We need to find the value of [tex]\( t \)[/tex] such that [tex]\( f(t) = 49 \)[/tex]. We'll set up the equation:

[tex]\[ \frac{7 - 8t}{7} = 49 \][/tex]

To clear the fraction, multiply both sides by 7:
[tex]\[ 7 - 8t = 343 \][/tex]

Next, solve for [tex]\( t \)[/tex]:

1. Subtract 7 from both sides:
[tex]\[ -8t = 343 - 7 \][/tex]
[tex]\[ -8t = 336 \][/tex]

2. Divide by -8:
[tex]\[ t = \frac{336}{-8} \][/tex]
[tex]\[ t = -42 \][/tex]

So, the value of [tex]\( t \)[/tex] that makes [tex]\( f(t) = 49 \)[/tex] is [tex]\( t = -42 \)[/tex].

### Summary

a. [tex]\( f(t) = \frac{7 - 8t}{7} \)[/tex]
b. [tex]\( f(-7) = 9 \)[/tex]
c. [tex]\( t = -42 \)[/tex]

I hope this helps! Let me know if you have any more questions.