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ryzh [129]
3 years ago
5

What is the recursive formula, if the sequence is 4, 7, 10, 13, ...?

Mathematics
2 answers:
elena-14-01-66 [18.8K]3 years ago
6 0

Answer:

4 + 3(n-1)

Step-by-step explanation:

Sequences start from n=1,2,3....

In this case

4 + 3(1-1)= 4

4 + 3(2-1) =7

And continue the pattern

tensa zangetsu [6.8K]3 years ago
4 0

Answer:

Step-by-step explanation:

In the arithmetic sequence \(t_1\), \(t_2\), \(t_3\), ..., \(t_n\), \(t_1=23\) and \(t_n= t_{n-1} - 3\) for each n > 1. What is the value of n when \(t_n = -4\)?

A. -1

B. 7

C. 10

D. 14

E. 20

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_________________

Best Regards,

E.

MGMAT 1 --> 530

MGMAT 2--> 640

MGMAT 3 ---> 610

GMAT ==> 730

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A toy company had an order of five boxes to ship unfortunately this was 17 pounds over the shipping weight limit per order of ea
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Answer:

28 pounds

Step-by-step explanation:

We have a total of 5 boxes, now we know that each box weighs 9 pounds, therefore:

Total weight = 5 * 9 = 45 pounds

Which means there are 45 pounds in total.

We are told that the limit is exceeded by 17 pounds. To find the limit weight, it is necessary to subtract the total weight and the excess:

45-17 = 28 pounds

Therefore, the maximum weight allowed per shipment is 28 pounds.

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Answer:

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The areas of two similar triangles are 72dm2 and 50dm2. The sum of their perimeters is 226dm. What is the perimeter of each of t
zloy xaker [14]

Answer:

103=  p small

plarge = 123

Step-by-step explanation:

We know that the the ratio of the areas is the scale factor squared/

Larger triangle over smaller triangle

72

----- = scale factor ^2

50

simplify by dividing by 2 on top and bottom

36

----- = scale factor ^2

25

Take the square root of each side

sqrt(36)

-------------- = sqrt(scale factor ^2)

sqrt(25)

6

-------------- = scale factor

5

The ratio of the perimeters is the scale factor

p large             6

-------------- = ----------------

 p small                5

Using cross products

6 p large = 5 p small

We know the sum is 226

p large + p small = 226

p large = 226 - p small

We have 2 equation and 2 unknowns

6 p large = 5 p small

Substitute for p large

6 (226 - p small) = 5 p small

1356 - 6 p small = 5 p small

Add 6 p small to each side

1356 = 11 p small

divide by 11

1356/11 =  p small

P large = 226-1356/11

p large = 2486/11-1356/11

plarge = 1130/11

The solution requires whole number answers so

1356/11 =  p small

123.27 which rounds to 123

plarge = 1130/11

plarge = 102.7272 = 103

6 0
3 years ago
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