Y=2000*

2200=2000*


<span>
so,
y at t=8 is 2000*(</span><span>

)^8
y = 2000(22/20)^2=2000*1.21=</span><span>2420 thats the answer i hope.</span>
The width is 6
Times 20 and 30 and it equals 600
So times 600 by 6 and 3600
I cant answer a problem that i cannot see
Answer:

Step-by-step explanation:
13 = 5 + 5 + 3





sum of the four consecutive integers is 238; The next step is to represent the four consecutive integers using the variable “ n “. Let n be the first integer. Since the four integers are consecutive, this means that the second integer is the first integer increased by 1 or {n + 1}.