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Sladkaya [172]
2 years ago
10

If f(1) = 2 and f(n) = f(n − 1)2 + 3 then find the value of f(3).

Mathematics
1 answer:
blondinia [14]2 years ago
3 0

Answer:

52

Step-by-step explanation:

f(n) is purely based on the previous value of f(n), or f(n-1), so we can start with f(1) and work our way up. We know f(1) = 2, so to find f(2), we plug f(1) into

f(n-1)²+3 to get

f(1)²+3 = 2²+3 = 4+3=7

Thus, f(2) =7. Similarly,

f(3) = f(3-1)²+3 = f(2)² + 3 -= 7² + 3= 52

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A rubber ball has a radius of about 2.86 in. Can the ball be packaged in a box shaped like a cube with a volume of 125 in3?
Vlada [557]

Answer:

I am not sure what the values needed to add in / etc, but here is the height of the box: 5

Step-by-step explanation:

A cube is a kind of rectangle where all the sides are the same. So to find volume, cube the length of any side. To find height, calculate the cube root of a cube's volume. For this example, the cube has a volume of 125. The cube root of 125 is 5. The height of the cube is 5.

(hopefully this is correct, have a nice day!)

6 0
2 years ago
Given that these simultaneous equations
Tatiana [17]

The value of k is(3\sqrt{ 10})/2

<h3>How to solve the simultaneous equation?</h3>

Given:

x-y=k.............(eq i)

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We would make y the subject formula in eq ii

2x²+y²-15= 0

2x² + y²= 15

y²= 15-2x²

y= \sqrt{15-2x^2}...........(eq iii)

Substitute the value of y into eq i

x-(\sqrt{15-2x^2}= k

x- (\sqrt{15} - 2x= k

k= (3\sqrt{ 10})/2

Read more about simultaneous equations here:

brainly.com/question/16863577

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3 0
1 year ago
Can someone please help me with my homework
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4 0
3 years ago
Find the quotient. (6 1/2 ÷ 3 2/3) ....1 17/22 23 5/6 21/3 2 3/4
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Step-by-step explanation:

7 0
2 years ago
What is the equation?​
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Answer:

  cost = 1.35n

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Step-by-step explanation:

Based on the numbers given, the cost is proportional to the number of songs downloaded. The constant of proportionality is the cost of one song: 1.35.

  cost = 1.35n

For 25 songs, ...

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__

The equation is ...

  cost = 1.35n

5 0
3 years ago
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