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DanielleElmas [232]
4 years ago
8

Give another name for AE

Mathematics
1 answer:
Gnoma [55]4 years ago
4 0

Answer:  \text{Line AE}=\overleftrightarrow{EA},\quad  \overleftrightarrow{DA}, \quad \overleftrightarrow{AD},\quad \overleftrightarrow{ED},\quad \overleftrightarrow{DE},\quad \text{line r}

               \text{Segment AE}=\overline{EA}

<u>Step-by-step explanation:</u>

The line can be named using ANY two points on the line or using the lower case letter.

The segment can be named by placing the letters in any order.

\overline{AE}=\overline{EA}

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Fantom [35]

Answer:

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3 years ago
In an arithmetic sequence, a17 = -40 and
Viktor [21]

Answer:

Tn = 2Tn-1 - Tn-2

Step-by-step explanation:

Before we can generate the recursive sequence, we need to find the nth term of the given sequence.

nth term of an AP is given as:

Tn = a+(n-1)d

If a17 = -40

T17 = a+(17-1)d = -40

a+16d = -40 ...(1)

If a28 = -73

T28 = a+(28-1)d = -73

a+27d = -73 ...(2)

Solving both equations simultaneously using elimination method.

Subtracting 1 from 2 we have:

27d - 16d = -73-(-40)

11d = -73+40

11d = -33

d = -3

Substituting d = -3 into 1

a+16(-3) = -40

a - 48 = -40

a = -40+48

a = 8

Given a = 8, d = -3, the nth term of the sequence will be

Tn = 8+(n-1) (-3)

Tn = 8+(-3n+3)

Tn = 8-3n+3

Tn = 11-3n

Given Tn = 11-3n and d = -3

Tn-1 = Tn - d... (3)

Tn-1 = 11-3n +3

Tn-1 = 14-3n

Tn-2 = Tn-2d...(4)

Tn-2 = 11-3n-2(-3)

Tn-2 = 11-3n+6

Tn-2 = 17-3n

From 3, d = Tn - Tn-1

From 4, d = (Tn - Tn-2)/2

Equating both common difference

(Tn - Tn-2)/2 = Tn - Tn-1

Tn - Tn-2 = 2(Tn - Tn-1)

Tn - Tn-2 = 2Tn-2Tn-1

2Tn-Tn = 2Tn-1 - Tn-2

Tn = 2Tn-1 - Tn-2

The recursive formula will be

Tn = 2Tn-1 - Tn-2

5 0
3 years ago
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Answer:

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Step-by-step explanation:

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4 years ago
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Answer:

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Step-by-step explanation:

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