How many years would it take for a $100 investment to become $800 at an APR of 12%, compounded quarterly?

soo this can be solve by using the formula:

f = p( 1 + i)^ n

where f is the future profit

p is the initial profit

i is the percent increase

n is the number of years

a . f = 30000 ( 1.05)^n

b. f = 30000 ( 1.05)^15

f = $62,368

тgє qυєѕтισи αѕкє∂: ιи тнє ѕєяιєѕ σя яєα¢тισиѕ тgαт σ¢¢υя ιи ρнσтσѕуитнєѕιѕ, ωнι¢н fσямѕ ωнєи ℓιgнт ѕтяιкєѕ α мσℓє¢υℓє σf ¢нℓσяσρнуℓℓ.

α. мσℓє¢υℓє σf ωαтєя

в.тнєє-¢αявσи ѕυgαя

¢. "єχ¢ιтє∂" єℓє¢тяσи

∂. иα∂ρн

му αиѕωєя: ιт ιѕ ¢. "єχ¢ιтє∂" єℓє¢тяσи

нσρє ι ¢συℓ∂ нєℓρ υ

a) 40 more; b) old factory:   10%, new factory:   16%; c) week 4

explanation:

a) at week 0, we see in the graph that the new factory produces 190 specialty items.  

using the function, substituting 0 in for w,

this is a difference of 230-190 = 40.

b) the function for the old factory is written in the form f(x) = a(1+r)ˣ, where a is the original amount, x is the amount of time and r is the growth rate.   the equation for the old factory is ; 230 is in place of a and is the original amount; w is the amount of time; which means 1+r = 1.1.   this means that r must be 0.1 or 10%.

for the new factory, we find the percent of increase to find the growth rate; the formula for this is

between the first two points, the amount of change is 220-190 = 30; this gives us

30/190 = 0.158 ≈ 0.16 = 16%

c) at week 0, the old factory has 230 while the new factory has 190.

at week 1, the old factory has 230(1.1) = 253 while the new factory has 220.

at week 2, the old factory has 230(1.1²) = 278.3 while the new factory has 252.

at week 3, the old factory has 230(1.1³) = 306.13 while the new factory has 290.

at week 4, the old factory has 230(1.1⁴) = 336.743 while the new factory has 337.   this is the first week that the new factory surpasses the old.