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pashok25 [27]
3 years ago
9

Complete the recursive formula of the arithmetic sequence -17, -8, 1, 10,... a(1)= a(n)=a(n-1)+

Mathematics
1 answer:
hichkok12 [17]3 years ago
8 0

To find a(1), simply look at the first term, which is -17. This gives a starting point for the recursive formula.

Then, we want to write the recursive formula as a(n) = a(n - 1) + c, where c is the number you add to a term to get the next term. You can see from the values given that you are adding 9 to each term to get the next one, so the recursive formula would be a(n) = a(n - 1) + 9.

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What is (1, -2) and (2, 2) as a linear equation
frutty [35]

Answer:

y=4

Step-by-step explanation:

y=2+2/2-1

y=4

8 0
2 years ago
In ΔRST, t = 4.1 inches, r = 7.1 inches and ∠S=45°. Find the length of s, to the nearest 10th of an inch.
dem82 [27]

Answer:

The length of s is 5.1 inches to the nearest tenth of an inch

Step-by-step explanation:

In Δ RST

∵ t is the opposite side to ∠T

∵ r is the opposite side to ∠R

∵ s is the opposite side to ∠S

→ To find s let us use the cosine rule

∴ s² = t² + r² - 2 × t × r × cos∠S

∵ t = 4.1 inches, r = 7.1 inches, and m∠S = 45°

→ Substitute them in the rule above

∴ s² = (4.1)² + (7.1)² - 2 × 4.1 × 7.1 × cos(45°)

∴ s² = 16.81 + 50.41 - 41.1677568

∴ s² = 26.0522432

→ Take √ for both sides

∴ s = 5.10413981

→ Round it to the nearest tenth

∴ s = 5.1 inches

∴ The length of s is 5.1 inches to the nearest tenth of an inch

3 0
3 years ago
Read 2 more answers
What’s the answer need help ?
Zielflug [23.3K]

Answer:

B

Step-by-step explanation:

Just move point D over 6 and down 10

4 0
3 years ago
Consider the quadratic function:<br> f(x) = x2 – 8x – 9
Bumek [7]

Solved in quadratic function
The answer :
X=(9,-1)
7 0
3 years ago
What is the volume of the solid?
Evgen [1.6K]

9514 1404 393

Answer:

  (9√3 -3π/2) ft^3 ≈ 10.88 ft^3

Step-by-step explanation:

The area of the hexagon is given by the formula ...

  A = (3/2)√3·s^2 . . . . for side length s

The area of the hexagonal face of this solid is ...

  A = (3/2)√3·(2 ft)^2 = 6√3 ft^2

__

The area of the circular hole in the hexagonal face is ...

  A = πr^2

The radius is half the diameter, so is r = (2 ft)/2 = 1 ft.

  A = π(1 ft)^2 = π ft^2

Then the area of the "solid" part of the face of the figure is ...

  A = (6√3 -π) ft^2

__

The volume is ...

  V = Bh . . . . . where B is the area of the base of the prism, and h is its height

  V = ((6√3 -π) ft^2)(3/2 ft) = (9√3 -3π/2) ft^3 ≈ 10.88 ft^3

8 0
3 years ago
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