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dmitriy555 [2]
3 years ago
5

Triangle A B C is shown. Lines are drawn from each point to the opposite side and intersect at point D. They form line segments

A G, B E, and C F.
In the diagram, which must be true for point D to be an orthocenter?

BE, CF, and AG are angle bisectors.
BE ⊥ AC, AG ⊥ BC, and CF ⊥ AB.
BE bisects AC, CF bisects AB, and AG bisects BC.
BE is a perpendicular bisector of AC, CF is a perpendicular bisector of AB, and AG is a perpendicular bisector of BC.
Mathematics
2 answers:
Mice21 [21]3 years ago
8 0

Answer:

B BE⊥AC, CF ⊥AB

Step-by-step explanation:

on Edge

Klio2033 [76]3 years ago
4 0

<em>b</em>

That is in fact the answer.

You know where I found it.

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Tell whether the sequence is arithmetic. Justify your answer. If the sequence is arithmetic, write a recursive
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Answer:

The explicit formula

a_n = 3 +0.25 (n-1)

The recursive formula

a_1 = 3                            for n=1

a_n = a_ {(n-1)} +0.25     if n>1

Step-by-step explanation:

If a sequence is arithmetical then the difference between any of its consecutive terms will be constant

3, 3.25, 3.5, 3.75,

3.25-3 = 0.25\\\\3.5-3.25 = 0.25\\\\3.75 -3.5 = 0.25

The difference between the consecutive terms remains constant so the sequence is arithmetic.

The explicit formula for an arithmetic sequence is:

a_n = a_1 + d (n-1)

Where d is the constant difference between the terms.

d = 0.25

a_1 is the first term of the sequence.

a_1 = 3

So

a_n = 3 +0.25 (n-1)

Finally, the recursive formula is:

a_1 = 3\\\\a_n = a_ {(n-1)} +0.25

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