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DaniilM [7]
3 years ago
5

Can someone help me on this play

Mathematics
2 answers:
jolli1 [7]3 years ago
8 0

Answer:

Q. 2-1/2

R. 1- 1/6

S. 2- 2/5

djyliett [7]3 years ago
3 0

Answer: 2 1/2   1 1/6   2 2/5

Step-by-step explanation:

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PLEASE HELP <br> 10TH GRADE GEOMETRY
FromTheMoon [43]

Answer:

perimeter = 4 \sqrt{26}

Step-by-step explanation:

Using the distance formula

AB = \sqrt{(-4 - 1)^{2} + (3 - 4)^{2} }

     = \sqrt{(-5)^{2} + (-1)^{2}  }

     = \sqrt{25 + 1}

     = \sqrt{26}

The perimeter of a square is 4 times the length of one side.

So,  perimeter = 4 \sqrt{26}

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How is 3×3×3×3×3×3×3 written in exponential form​
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Answer:

7

Step-by-step explanation:

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3 years ago
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Can someone please help me with #1,2,3,4 and 5 using a motion detector?
Olin [163]

Problem 5

Recall that speed = distance/time. A change in distance over a change in time also represents the slope of the line.

The slope of any given piece represents the walker's speed. Negative slopes mean that the walker is coming closer to the device, while positive slopes mean the walker is moving away.

For instance, a slope of m = -2 means the walker is moving at 2 mph and is moving toward the motion detector. This is just an example of course.

The graph given here starts off with a negative sloped portion. It goes downhill as we move to the right because time always progresses in this fashion (i.e. time is always increasing). So this portion represents the walker moving closer to the device at some unknown speed. We can't calculate the speed because your teacher didn't give you any numbers. So you can make up your own or just give the general sense of what's going on.

For the middle portion, the flat horizontal line means that slope = rise/run = 0/B = 0. Where B is some positive number to indicate a run. The slope being 0 tells us that the speed is 0, which in turn means the person isn't moving at all during this stretch of time.

After being motionless for some unknown amount of time, the walker then moves away from the motion detector. This is because this third portion has a positive slope now.

To recap, we have this summary of motion (or lack thereof)

  1. The person moves closer to the motion detector
  2. The person is motionless
  3. The person moves away from the motion detector

Each section has an unknown duration and unknown speed (other than the speed of 0 mentioned). The sections are numbered to indicate the proper order in time in which they happened.

============================================================

Problem 6

It's the same idea as earlier. The person starts off walking away from the motion detector. Then at some point, they turn around and walk back toward the detector until they reach the device itself. How do we know the person reaches the device? Because of the x intercept. This is when y = 0 to indicate the distance to the device is 0 units. The x intercept is where the graph crosses the x axis.

Answer:

  1. The person moves further away from the detector
  2. The person turns around and moves closer to the detector until reaching the device

Once again, we have no idea about speeds or time durations. This time a speed of 0 isn't involved, so the person is never motionless (maybe perhaps for a brief split instant, but that's about it).

============================================================

Problem 7

There are no increasing sections this time. The person never walks away from the motion device at any moment. This is due to the lack of positive sloped sections.

S/he starts at some distance away and walks closer to the device for the first section of time. Then they pause and stay motionless for a while similar to what was discussed in problem 5. Then the person resumes motion to walk to the motion detector until they reach the device. We know they reach it because of the x intercept.

Answer:

  1. The person starts at some distance away and walks closer to the device.
  2. The person pauses for some duration.
  3. The person resumes motion toward the motion detector.

============================================================

Problem 8

This time we're dealing with curved sections, but the idea is pretty much the same. Any time we go uphill, the slope is positive (so to speak) and it represents moments the person is walking away. Any time we go downhill, the person is walking toward the item. Always move to the right because time is always increasing in this fashion.

The smooth curve motion indicates that the person's speed is changing at a more smooth rate, as compared to the sharp pointed jarring changes in speed compared to the other graphs. This is likely a more realistic graph for someone's motion. Someone is very unlikely to change speeds very quickly in such a short instant in time.

Answer:

  1. The person walks away from the device
  2. They walk toward the device
  3. They change their mind to move away
  4. They change their mind yet again to come back to the motion detector

============================================================

Problem 9

We have four separate disconnected sections that are all flat. This indicates that we have four people at different (unknown) distances away from the motion detector. None of these people are moving.

The unfortunate thing is that this graph is a bit misleading. This graph has the first section end rather than go on forever. The first person (or any other person for that matter) doesn't stop being detected and they aren't magically teleported away. However, the graph is set up like this to avoid failing the vertical line test.

3 0
2 years ago
find the point-slope equation of the line perpendicular to the line y=2x+5 passing through the point (1,4). (please include solu
pishuonlain [190]
First off, let's keep in mind that perpendicular lines have negative reciprocal slopes, hmmm so what's the slope of y = 2x + 5?

well, notice, that equation is in slope-intercept form, thus   \bf y=\stackrel{slope}{2}x+5.

\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}}&#10;{2\implies \stackrel{slope}{\cfrac{2}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{2}}\qquad \stackrel{negative~reciprocal}{-\cfrac{1}{2}}}

so, we're really looking for the equation of a line whose slope is -1/2 and runs through 1,4.

\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad \qquad \qquad &#10;slope =  m\implies -\cfrac{1}{2}&#10;\\\\\\&#10;% point-slope intercept&#10;\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-4=-\cfrac{1}{2}(x-1)
7 0
3 years ago
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