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taurus [48]
4 years ago
11

Show that it is not possible for the lengths of the

Mathematics
1 answer:
xeze [42]4 years ago
7 0
1. Check the picture below.

2. By the well known intersecting secants property: AO*OC=BO*OD

3. Since the segments are assumed to be consecutive integers, AO, OC cannot be the smaller, and neither can AO<BO and OC<OD. 
The only possibility is that the multiplication of the smallest to the largest is equal to the multiplication of the middle 2.

4. For example : 4*5=2*10, we must have 4>2 but 5<10

Here the ordering is 2<4<5<10, similarly assume BO<AO<CO<DO

5. since the segments are of consecutive integers length, let them be 

BO=n, AO=n+1, CO=n+2, DO=n+3

6. AO*CO=BO*DO

so n(n+3)=(n+1)(n+2)
     n^{2}+3n=  n^{2}+3n+3
     so 0=3 which makes no sense, so no solution with the assumed conditions.

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the sum of three consecutive even integers is 36. Let X represent the first integer, x + 2 represent the second integer, and x +
ozzi

Answer:

see explanation

Step-by-step explanation:

Sum the 3 consecutive integers and equate to 36

x + x + 2 + x + 4 = 36

3x + 6 = 36 ( subtract 6 from both sides )

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6 0
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MN is the midsegment of the trapezoid ABCD. What is the length of segment AB?
vladimir1956 [14]

Dear Student,

Answer to your query is provided below:

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Explanation to answer is provided by attaching image.

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