Suppose a farmer has six cows and each cow has six calves the first year, they all increase in the same propotion, to the end of eight years, how many calves will the farmer have? ( mention the steps and fifty points and brainliest will be given to the one who answers this question)

Hi!

To solve this question you need to solve 7^8 x 6

7^8 = 5,764,801

5,764,801 x 6 = 34,588,806

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)  

∈3,600

step-by-step explanation: