The answer would be C. hope that helps out :)
1st let's calculate the decreasing rate & let V₁ be the initial value & V₂ the final's
we know that V₂=V₁.e^(r,t) where r=rate & t-time (& e=2.718)
After t= 2 years we can write the following formula
2350,000=240,000.e^(2r)==> 235,000/240000 = e^(2r) =>47/48=e^(2r)
ln(47/48)=2rlne==> ln(47/48)=2rlne=2r (since lne =1)
r= ln(47/48)/2==>r=-0.0210534/2 =-0.01052 ==> (r=-0.01052)
1) Determine when the value of the home will be 90% of its original value.
90% of 240000 =216,000
Now let's apply the formula
216,000=240,000,e^(-0.01052t), the unknown is t. Solving it by logarithm it will give t=10 years
1.a) Would the equation be set up like so: V=240e^.09t? NON, in any case if you solve it will find t=1 year
2)Determine the rate at which the value of the home is decreasing one year after : Already calculated above :(r=-0.01052)
3)The relative rate of change : it's r = -0.01052
Answer:
a,b,d,e for sho on engunity
3/4 - 1/8 = 6/8 - 1/8 = 5/8...so 5/8 = 10 gallons
(5/8) / 10 = 1 / x
cross multiply
(5/8)(x) = (10)(1)
5/8x = 10
x = 10 * 8/5
x = 80/5 = 16 <== so it would take 16 gallons to fill the tank
94 divided by 33 is 2.8484848484848
