Lerna's Hydra is a mythological character that appears in some stories such as the 12 Tasks of Hercules. The Hydra was a one-headed monster but when it is cut off, 2 more heads grow in its place. If a hero tried to conquer it by cutting off all of its heads every day, how many heads would the Hydra have on the third day? And at the end of 10 days of trying to kill it? *

The number of heads that would the Hydra have on the third day is 8

And, at the end of 10 days of trying to kill it is 1024.

a = 1

r = 2 by 1 = 2  

T4

= 8

So the hydra will have 8 heads on the third day.

Now  

To know the number of heads at the end of 10 days, we will determine the 11th term of the progression

= 1024

Learn more: link

Step-by-step explanation:

Initially the monster is one-headed, when it is cut off, 2 more heads grow in its place, when the two heads are cut off, 4 will grow in its place. This will go on an on and the amount of heads that grows each day will form a geometric sequance as shown;

1, 2, 4...

The nth term of an arithmetic sequence is expresses as;

where;

a is the first term

r is the common ratio

n is the number of terms

To get the number of heads that the Hydra have on the third day, we will have to find the fourth term of the sequence since the first cutting was done the next day (second day).

from the sequence above;

a = 1

r = 2/1 = 4/2 = 2

T4 = 1(2)^{4-1}

T4 = 2^3

T4 = 8

Hence the hydra will have 8 heads on the third day.

To know the number of heads at the end of 10 days, we will find the 11th term of the progression

T11 = 1(2)^{11-1}

T4 = 2^10

T4 = 1024

He will have 1024 heads after 10 days of trying to kill the monster.

the answer is (4,3)

since the rule for r0, 90º is (-y,x) and it was already applied to form the image, all you have to do is reverse the answer using this formula.