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3 years ago

What 1/3 of an hour equivalent to

Mathematics

2 answers:

myrzilka [38]3 years ago

8 0

Answer:

$\frac{1}{3}$ of a hour = 20 minutes

Step-by-step explanation:

Hope this helps .^.

ehidna [41]3 years ago

5 0

Answer:20 minutes

Step-by-step explanation: The reason why its 20 is because, 1/3 0f 60 mintues is 20 mintues. 60/3.

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Josh thinks he can find examples of prime numbers that he can write as a product of two points

LiRa [457]

It would count on the graph mainly (or table). But yes it can be possible.

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3 years ago

On a coordinate graph, points F (4,1), G (4,-1), H (-7,1) and J (-7,-1) are the vertices of a rectangle. What is the area of the

rosijanka [135]

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3 years ago

Match each function to its corresponding sequence.

givi [52]

The sequence and their functions are:

f(n) = f(n - 1) + 7 ⇒ 3, 10, 17, 24, 31.....
f(n) = f(n - 1) + 4n - 2 ⇒ 3, 9, 19, 33, 51.....
f(n) = 2[f(n - 1)] ⇒ 3, 6, 12, 24, 48.....

<h3>How to match the functions?</h3>

To do this, we simply set values for n and calculate the function values using the current value of n.

So, we have:

Function 1: f(n) = f(n - 1) + 6n - 5 where f(1) = 3

Let n = 2

f(2) = f(1) + 6(2) - 5 = 3 + 12 - 5 = 10

Let n = 3

f(3) = f(2) + 6(3) - 5 = 10 + 18 - 5 = 23

None of the sequence follows the pattern 3, 10, 23....

Function 2: f(n) = f(n - 1) + 2n - 1 where f(1) = 3

Let n = 2

f(2) = f(1) + 2(2) - 1 = 3 + 4 - 1 = 6

Let n = 3

f(3) = f(2) + 2(3) - 1 = 6 + 6 - 1 = 11

None of the sequence follows the pattern 3, 6, 11....

Function 3: f(n) = f(n - 1) + 6 where f(1) = 3

Let n = 2

f(2) = f(1) + 6 = 3 + 6 = 9

Let n = 3

f(3) = f(2) + 6 = 9 + 6 = 15

None of the sequence follows the pattern 3, 9, 15....

Function 4: f(n) = f(n - 1) + 7 where f(1) = 3

Let n = 2

f(2) = f(1) + 7 = 3 + 7 = 10

Let n = 3, 4 and 5

f(3) = f(2) + 7 = 10 + 7 = 17

f(4) = f(3) + 7 = 17 + 7 = 24

f(5) = f(4) + 7 = 24 + 7 = 31

So, we have:

f(n) = f(n - 1) + 7 ⇒ 3, 10, 17, 24, 31.....

Function 5: f(n) = f(n - 1) + 4n - 2 where f(1) = 3

Let n = 2

f(2) = f(1) + 4(2) - 2 = 3 + 8 - 2 = 9

Let n = 3, 4 and 5

f(3) = f(2) + 4(3) - 2 = 9 + 12 - 2 = 19

f(4) = f(3) + 4(4) - 2 = 19 + 16 - 2 = 33

f(5) = f(4) + 4(5) - 2 = 33 + 20 - 2 = 51

So, we have:

f(n) = f(n - 1) + 4n - 2 ⇒ 3, 9, 19, 33, 51.....

Function 6: f(n) = 2[f(n - 1)] where f(1) = 3

Let n = 2

f(2) = 2 * f(1) = 2 * 3 = 6

Let n = 3, 4 and 5

f(3) = 2 * f(2) = 2 * 6 = 12

f(4) = 2 * f(3) = 2 * 12 = 24

f(5) = 2 * f(4) = 2 * 24 = 48

So, we have:

f(n) = 2[f(n - 1)] ⇒ 3, 6, 12, 24, 48.....