#### rachelmaddie

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GI = 5b + 2, HI = 4b – 5, HI = 3

My work:

4b - 5 = 3

5 + 3 = 8

4b = 8

b = 2

GI = 5b + 2, and since b = 2

GI = 5(2) + 2

GI = 12

I don’t know how to get to GH.

- Thread starter rachelmaddie
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GI = 5b + 2, HI = 4b – 5, HI = 3

My work:

4b - 5 = 3

5 + 3 = 8

4b = 8

b = 2

GI = 5b + 2, and since b = 2

GI = 5(2) + 2

GI = 12

I don’t know how to get to GH.

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- 19,982

Is there a drawing showing G, H & I are collinear (in one st. line)?

GI = 5b + 2, HI = 4b – 5, HI = 3

My work:

4b - 5 = 3

5 + 3 = 8

4b = 8

b = 2

GI = 5b + 2, and since b = 2

GI = 5(2) + 2

GI = 12

I don’t know how to get to GH.

Please double check to make sure you copied the problem down correctly. TakenFind the value of the variable and GH if H is between G and I.

GI = 5b + 2, HI = 4b – 5, HI = 3

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No drawing.Is there a drawing showing G, H & I are collinear (in one st. line)?

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This is the exact problem..

Please double check to make sure you copied the problem down correctly. Takenexactlyas you wrote it here, the question makes no sense and is unanswerable. The line segment from the point G to the point I (denoted GI) is given by y = 12. This is a horizontal line. Likewise, the line segment from the point H to the point I (denoted HI) is given by y = 3. This is also a horizontal line. These two line segments must share one end point (the point I), but how can this be when the two lines are both horizontal and thus parallel to one another?

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\(\displaystyle \underbrace {G\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_H}_9\underbrace {\_\_\_\_\_\_\_\_\_\_I}_3\)

GI = 5b + 2, HI = 4b – 5, HI = 3

My work:

4b - 5 = 3

5 + 3 = 8

4b = 8

b = 2

GI = 5b + 2, and since b = 2

GI = 5(2) + 2

GI = 12

I don’t know how to get to GH.

The above is what I get from \(\displaystyle b=2,~GI=12,~\&~HI=3\)

Last edited:

At crux here is that the OP's notation apparently uses \(GI\) to refer to the

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Can you please explain to me how you got GH.\(\displaystyle \underbrace {G\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_H}_9\underbrace {\_\_\_\_\_\_\_\_\_\_I}_3\)

The above is what I get from \(\displaystyle b=2,~GI=12,~\&~HI=3\)

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\(\displaystyle 12-3=9\) With this posting you have convinced me that this material is really over your head.Can you please explain to me how you got GH.

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Mr. Khan the statement that point \(\displaystyle B\) is between points \(\displaystyle A~\&~C\) written \(\displaystyle A-B-C\) means that \(\displaystyle A,~B,~\&~C\) are colinear and \(\displaystyle AB+BC=AC\).Is there a drawing showing G, H & I are collinear (in one st. line)?

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Well, the problem is that I never learned it properly because the content for measuring segments in my textbook (which is the only form of learning) was actually different from what I had to apply to the assessments in the online course.\(\displaystyle 12-3=9\) With this posting you have convinced me that this material is really over your head.

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Well I feel sorry for you. I gave up teaching because of on-line courses. I was trained in the Moore-method of teaching.Well, the problem is that I never learned it properly because the content for measuring segments in my textbook (which is the only form of learning) was actually different from what I had to apply to the question assessments in the online course.

I just could not figure out how to teach a graduate mathematics course online.

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Well I feel sorry for you. I gave up teaching because of on-line courses. I was trained in the Moore-method of teaching.

I just could not figure out how to teach a graduate mathematics course online.

Well, the problem is that I never learned it properly because the content for measuring segments in my textbook (which is the only form of learning) was actually different from what I had to apply to the assessments in the online course.

[/QUOTE

I have spent a sufficient amount of time taking notes and reviewing all the lesson material (really the basic formulas for measuring segments) and it was nothing alligned to these more complex questions.

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I don’t understand how students can learn like that.