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sasho [114]
3 years ago
7

I WILL GIVE BRAINLIEST TO WHOEVER IS CORRECT

Mathematics
1 answer:
labwork [276]3 years ago
6 0
The second one should by right
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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
Read 2 more answers
Items for a fundraiser are packaged in small boxes shaped like rectangular prisms that are inches long, inches wide, and 8 inche
allochka39001 [22]

Answer:

- The number of small boxes that will fill the large box 1 = 64

- The number of small boxes that will fill the large box 2 = 56

Step-by-step explanation:

Complete Question

Items for a fundraiser are packaged in small boxes shaped like rectangular prisms that are 4.5 inches long, 4.5 inches wide, and 8 inches tall. To transport the items to an event, you want to know how many of the small boxes will fit in larger boxes. The larger boxes are available in two sizes. Large Box 1 is 24.25 inches long, 18 inches wide, and 24 inches tall. Large Box 2 is 20.5 inches long, 18.5 inches wide, and 24 inches tall. Both the small and large boxes must remain upright.

Solution

To know how many of the small boxes will fit in larger boxes, we need to obtain the volumes of the small box, large box 1 and large box 2.

Volume of a cuboid = L × W × H

For the small box,

Length = L = 4.5 inches

Width = W = 4.5 inches

Height = H = 8 inches

Volume of the small box = 4.5 × 4.5 × 8 = 162 in³

For large box 1,

Length = L = 24.25 inches

Width = W = 18 inches

Height = H = 24 inches

Volume of the large box 1 = 24.25 × 18 × 24 = 10,476 in³

For large box 2

Length = L = 20.5 inches

Width = W = 18.5 inches

Height = H = 24 inches

Volume of the large box 2 = 20.5 × 18.5 × 24 = 9,102 in³

The number of small boxes that'll fill the large box 1 = (10,476/162) = 64.667 = 64 small boxes (rounded down because the fraction cannot be forced into the large box 1.

The number of small boxes that will fill the large box 2 = (9,102/162) = 56.185 = 56 small boxes.

Hope this Helps!!!

7 0
3 years ago
What figure is comprised of two rays that share a common end point called a vertex?
KIM [24]
This would be called an ANGLE.

Hopefully this helps!
5 0
3 years ago
Find an explicit rule for the nth term of a geometric sequence where the second and fifth terms are -6 and 162, respectively. (2
Stells [14]

Answer: C is the answer

5 0
3 years ago
Here are the questions:
yaroslaw [1]

Answer:

Q1) 0.46 Q2)0.2333

Step-by-step explanation:

Q1) P=(4+12+7)/50 =0.46

Q2) P=7/(4+7+10+9)

6 0
3 years ago
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