Question

score on last attempt 0 out of 3
score in gradebook: 0 out of 3

a. in the above problem we found that when the time changed by 7 minutes away from 4:04 pm, the car’s distance changed by \underline{\hspace{1em}} miles away from 29 miles.
what new point exists in the relationship? $(x, y) = \underline{\hspace{1em}}$ \\( \boldsymbol{\text{(span class=\aminput\ javal notation. (span))}}
\text{try again. after traveling for an additional 7 minutes since 4:04pm how long have you traveled for since 4:00pm. similarly after traveling an additional 8.75? after mile marker 29 how far have you traveled since mile marker 24?}

b. complete the following:
when given a point in a relationship, $(x_1, y_1)$, and the constant rate of change, $m$, with which the two quantities’ values $x$ and $y$ are changing, the change in $y$ away from $y_1$ can be represented as
$y - y_1 = \underline{\hspace{1em}}$ \\( \boldsymbol{\text{hint preview syntax error}}

\text{submit}

question 5 points possible: 3
unlimited attempts.
score on last attempt: 0. score in gradebook: 0

Explanation:

Part b Solution:

Step 1: Recall the slope formula concept

The constant rate of change \( m \) (slope) is given by \( m=\frac{y - y_1}{x - x_1} \), where \( (x_1,y_1) \) is a point and \( (x,y) \) is another point. Rearranging this formula to solve for the change in \( y \) (i.e., \( y - y_1 \)) gives the relationship between the change in \( x \) (i.e., \( x - x_1 \)) and the rate of change \( m \).

Step 2: Derive the formula for \( y - y_1 \)

Starting with \( m=\frac{y - y_1}{x - x_1} \), multiply both sides by \( (x - x_1) \) to isolate \( y - y_1 \). So we get \( y - y_1=m(x - x_1) \).

Answer:

\( m(x - x_1) \)

Note for Part a:

Since the original problem (before the screenshot) is not fully visible (we don't have the initial point or the rate of change), we can't solve part a without more information. But if we assume the initial time at 4:04 pm is \( x_1 \) (in minutes since some start time) and initial distance \( y_1 = 29 \) miles, and the change in time is 7 minutes (so new time \( x=x_1 + 7 \)) and change in distance is \( \Delta y \) (let's say we found \( \Delta y=8.75 \) from the hint about additional 8.75 miles), then the new point would be \( (x_1 + 7,29 + 8.75)=(x_1 + 7,37.75) \). But without the initial \( x_1 \) (like time since 4:00 pm), we can't give a numerical \( x \)-value. If we assume 4:04 pm is 4 minutes after 4:00 pm, then \( x_1 = 4 \), so new \( x=4 + 7=11 \) and new \( y=29 + 8.75 = 37.75 \), so the point would be \( (11,37.75) \). But this is based on assumptions from the hint. If you can provide the initial problem details (like initial time and distance, rate of change), we can solve part a precisely.