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Leviafan [203]
3 years ago
9

How can you determine the difference between an arithmetic and geometric sequence if you are given the first 4 terms of the sequ

ence?
Mathematics
2 answers:
Bess [88]3 years ago
3 0
Compute successive differences of the terms.

If they are all the same, the sequence is arithmetic and the common difference is the difference you have found.

If successive pairs of differences have the same ratio, the sequence is geometric and the common ratio is the ratio you have determined.


Example of arithmetic sequence:
  1, 3, 5, 7
Successive differences are 3-1 = 2, 5-3 = 2, 7-5 = 2. All the differences are 2, which is the common difference of the sequence.

Example of geometric sequence:
  1, -3, 9, -27
Successive differences are -3-1 = -4, 9-(-3) = 12, -27-9 = -36. These are not the same, so the sequence is not arithmetic. Ratios of successive pairs of differences are 12/-4 = -3, -36/12 = -3. These are the same, so the sequence is geometric with common ratio -3.
Amanda [17]3 years ago
3 0

Sample response: First identify the pattern of the sequence. If there is a common difference between the terms, it is an arithmetic sequence. If there is a common ratio between the terms, it is a geometric sequence.

Answer on edge nuity : )

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Please I need help ASAP<br>pls explain your answer ​
USPshnik [31]

Answer:

47.62 cm

Step-by-step explanation:

ABCD is a rhombus. Each side measures 15 cm.

m<A = 60°.

Here are some properties of a rhombus:

Opposite angles of a rhombus are congruent.

m<C = m<A = 60°

m<ADC = m<ABC

The sum of the measures of the interior angles of a quadrilateral is 360°.

The diagonals are perpendicular bisectors of each other.

m<A + m<C + m<ADC + m<ABC = 360°

60° + 60° + 2m<ADC = 360°

2m<ADC = 240°

m<ADC = 120°

Each diagonal of a rhombus bisects a pair of congruent, opposite angles.

m<ADB + m<CDB = m<ADC

m<ADB = m<CDB

m<ADB + m<ADB = 120°

m<ADB = 60°

Triangle ABD had two angles, <A and <ADB, that measure 60°. Therefore, the third angle, <ABD also measures 60°. That makes triangle ABD an equilateral triangle.

Draw segment AT.

T is the midpoint of BD. The segment AT is the perpendicular bisector of segment BD. Triangle ATD is a 30-60-90 triangle.

Using the ratio of the lengths of the sides of a 30-60-90 triangle, we can calculate the length of AT which is the radius of circle A.

TD : AT : AD

1    : √3 :   2

AD = 15 cm

TD = AD/2

AT = √3 × TD

AT = (AD√3)/2

AT = 7.5√3 = 12.99

AQ = 12.99

perimeter = QD + PB + m(arc)PTQ + BC + CD

perimeter = (15 - 12.99) + (15 - 12.99) + 60°/360° × 2π(12.99) + 15 + 15

perimeter = 47.62

Answer: perimeter = 47.62 cm

7 0
2 years ago
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