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Amiraneli [1.4K]
3 years ago
6

⦁ Scott works as a delivery person for a shipping company. The graph shows a linear model for his delivery times on different da

ys.
Questions are on the next page.
⦁ Graph is on the previous page

⦁ Identify two points on the line and write the coordinate pairs below.

⦁ Use your two points to find the slope, and make sure to show your work.

⦁ Use your slope from Part B and find the equation of the line in point-slope form. Show your work and work through the steps to transform the equation from point-slope form into slope-intercept form.

⦁ Based on the equation you found in Part C, predict how long it initially took Scott to deliver his packages. Approximately how much did his delivery time decrease per day?

Mathematics
1 answer:
BARSIC [14]3 years ago
6 0

Answer:

  • (3, 21), (6, 12)
  • -3
  • y -21 = -3(x -3)  ⇒  y = -3x +30
  • 30 minutes; 3 minutes

Step-by-step explanation:

a) The two marked points have coordinates (days, minutes) = (3, 21) and (6, 12).

__

b) The slope is the vertical change (rise) divided by the horizontal change (run). For these two points, that is ...

  rise/run = (12-21)/(6-3) = -9/3 = -3 = slope

__

c) The point-slope form of the equation for a line is ...

  y -k = m(x -h) . . . . . . for slope m through point (h, k)

Using the first point and the slope we found, this is ...

  y -21 = -3(x -3) . . . . . point-slope form

Add 21 and eliminate parentheses:

  y = -3x +9 +21

  y = -3x +30 . . . . . . . . collect terms; slope-intercept form

__

d) The initial time is the time corresponding to day 0, the value of y when x=0. It is 30 minutes.

The slope in part (b) tells us the time decreases by 3 minutes per day.

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\huge\fbox\red{a}\huge\fbox\orange{n} \huge\fbox\pink{s}\huge\fbox\green{w} \huge\fbox\blue{e}\huge\fbox\purple{r}\

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Whoever checks my work and tell what I did wrong and the right answer will
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Answer:

See below

Step-by-step explanation:

I believe that you only had to do letters F, H, and J. In that case, let's go over each one!

F: For isolatingd_{1}, we need to get rid of the 1/2 first. Let's multiply each side by 2:

2*m = 2*\frac{1}{2}(d_{1}+  d_{2}) \\2m = d_{1}+  d_{2}

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G: Let's isolate the vw^2 on one side by subtracting y from each side:

vw^{2} +y-y=x-y\\vw^{2}=x-y

Let's now divide each side by v, then put each side under a square root to get our final answer:

\frac{vw^2}{v} = \frac{x-y}{v}\\  w^2 =  \frac{x-y}{v}\\\sqrt{w^2} = \sqrt{\frac{x-y}{v}} \\w=\sqrt{\frac{x-y}{v}}

Orange is correct! Again, let's solve the problem underneath:

w=w=\sqrt{\frac{38-(-7)}{5}} = \sqrt{\frac{38+7}{5}} = \sqrt{\frac{45}{5}}=\sqrt{9}=3

H: This one has some stuff that we haven't worked with quite yet (like terms), but our approach is the same: isolate c on one side of the equation.

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Purple is correct! Let's solve the problem:

c = 5(\frac{16}{5})+7 = 16+7 = 23

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