The formula for calculating compound interest with yearly contributions is:
Balance = X*(1 + Y)^n + Z((1 + Y)^(n + 1) - (1 + Y)/Y)
where the balance is the money earned after n years invested
Y is the interest rate as a fraction
Z is the yearly contribution
X is the starting investment
Therefore the calculation for this example is:
Balance = 1200*(1 + 0.05)^48 + 1200((1.05)^49 - (1.05)/05)
= $249,393.5
Answer:
Option a. convergent: A
Step-by-step explanation:
a0=3/2
a1=9/8
a2=27/32
a1/a0=(9/8)/(3/2)=(9/8)*(2/3)→a1/a0=3/4
a2/a1=(27/32)/(9/8)=(27/32)*(8/9)→a2/a1=3/4
r=a1/a0=a2/a1→r=3/4
The abosute value of r= !r! = !3/4! = 3/4 = 0.75<1, the series is convergent
Answer:
![\boxed{-3xy^{2}\sqrt [3] {2x^{2}}}](https://tex.z-dn.net/?f=%5Cboxed%7B-3xy%5E%7B2%7D%5Csqrt%20%5B3%5D%20%7B2x%5E%7B2%7D%7D%7D)
Step-by-step explanation:
Your expression is
![\sqrt [3] {-54x^{5}y^{6}}](https://tex.z-dn.net/?f=%5Csqrt%20%5B3%5D%20%7B-54x%5E%7B5%7Dy%5E%7B6%7D%7D)
Here's how I would simplify it.
![\begin{array}{rcll}\sqrt [3] {-54x^{5}y^{6}} & = & \sqrt [3] {(-1)^{3}\times 2 \times 27 \times x^{2} \times x^{3} \times y^{6}} & \text{Factored the cubes}\\& = & \sqrt [3] {(-1)^{3} \times 3^{3}\times x^{3} \times y^{6}\times 2 \times x^{2}} & \text{Grouped the cubes}\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcll%7D%5Csqrt%20%5B3%5D%20%7B-54x%5E%7B5%7Dy%5E%7B6%7D%7D%20%26%20%3D%20%26%20%5Csqrt%20%5B3%5D%20%7B%28-1%29%5E%7B3%7D%5Ctimes%202%20%5Ctimes%2027%20%5Ctimes%20x%5E%7B2%7D%20%5Ctimes%20x%5E%7B3%7D%20%5Ctimes%20y%5E%7B6%7D%7D%20%26%20%5Ctext%7BFactored%20the%20cubes%7D%5C%5C%26%20%3D%20%26%20%5Csqrt%20%5B3%5D%20%7B%28-1%29%5E%7B3%7D%20%5Ctimes%203%5E%7B3%7D%5Ctimes%20x%5E%7B3%7D%20%5Ctimes%20y%5E%7B6%7D%5Ctimes%202%20%5Ctimes%20x%5E%7B2%7D%7D%20%26%20%5Ctext%7BGrouped%20the%20cubes%7D%5C%5C%5Cend%7Barray%7D)
![\begin{array}{rcll}& = & \sqrt [3] {(-1)^{3} \times {3^{3}\times x^{3} \times y^{6}}} \times\sqrt [3] { 2 \times x^{2}} & \text{Separated the cubes}\\&=& \mathbf{-3xy^{2}\sqrt [3] {2x^{2}}} & \text{Took cube roots}\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcll%7D%26%20%3D%20%26%20%5Csqrt%20%5B3%5D%20%7B%28-1%29%5E%7B3%7D%20%5Ctimes%20%7B3%5E%7B3%7D%5Ctimes%20x%5E%7B3%7D%20%5Ctimes%20y%5E%7B6%7D%7D%7D%20%5Ctimes%5Csqrt%20%5B3%5D%20%7B%202%20%5Ctimes%20x%5E%7B2%7D%7D%20%26%20%5Ctext%7BSeparated%20the%20cubes%7D%5C%5C%26%3D%26%20%5Cmathbf%7B-3xy%5E%7B2%7D%5Csqrt%20%5B3%5D%20%7B2x%5E%7B2%7D%7D%7D%20%26%20%5Ctext%7BTook%20cube%20roots%7D%5C%5C%5Cend%7Barray%7D)
![\text{The simplified expression is $\boxed{\mathbf{-3xy^{2}\sqrt [3] {2x^{2}}}}$}](https://tex.z-dn.net/?f=%5Ctext%7BThe%20simplified%20expression%20is%20%24%5Cboxed%7B%5Cmathbf%7B-3xy%5E%7B2%7D%5Csqrt%20%5B3%5D%20%7B2x%5E%7B2%7D%7D%7D%7D%24%7D)
Answer:
3 months
Step-by-step explanation:
t^3 - 10t^2 + 33t - 36
3t^2 -20t + 33 = 0
(3t-11)(t-3) = 0
t = 3 or t=11/3
The correct answer to this problem is be because you have to multiply the 4 outside of the parentheses with the 4 inside of the parentheses which gives you 16X and then afterwards you have to do the same with the 6 with so since 4•6=24 you can automatically eliminate answer choice D and then you add up your 3x to your 16x which will be 19x so it would be B. 19x+24