Answer:
The worth of the car after 6 years is £2,134.82
Step-by-step explanation:
The amount at which Dan buys the car, PV = £2200
The rate at which the car depreciates, r = -0.5%
The car's worth, 'FV', in 6 years is given as follows;
Where;
r = The depreciation rate (negative) = -0.5%
FV = The future value of the asset
PV = The present value pf the asset = £2200
n = The number of years (depreciating) = 6
By plugging in the values, we get;
The amount the car will be worth which is its future value, FV after 6 years is FV ≈ £2,134.82 (after rounding to the nearest penny (hundredth))
Four times a number would equal 4n
Complete Question
If $12000 is invested in an account in which the interest earned is continuously compounded at a rate of 2.5% for 3 years
Answer:
$ 12,934.61
Step-by-step explanation:
The formula for Compound Interest Compounded continuously is given as:
A = Pe^rt
A = Amount after t years
r = Interest rate = 2.5%
t = Time after t years = 3
P = Principal = Initial amount invested = $12,000
First, convert R percent to r a decimal
r = R/100
r = 2.5%/100
r = 0.025 per year,
Then, solve our equation for A
A = Pe^rt
A = 12,000 × e^(0.025 × 3)
A = $ 12,934.61
The total amount from compound interest on an original principal of $12,000.00 at a rate of 2.5% per year compounded continuously over 3 years is $ 12,934.61.
Answer:
(x^4-8)^45 /180 +c
Step-by-step explanation:
If u=x^4-8, then du=(4x^3-0)dx or du=4x^3 dx by power and constant rule.
If du=4x^3 dx, then du/4=x^3 dx. I just divided both sides by 4.
Now we are ready to make substitutions into our integral.
Int(x^3 (x^4-8)^44 dx)
Int(((x^4-8)^44 x^3 dx)
Int(u^44 du/4)
1/4 Int(u^44 dul
1/4 × (u^45 / 45 )+c
Put back in terms of x:
1/4 × (x^4-8)^45/45 +c
We could multiply those fractions
(x^4-8)^45 /180 +c
