The formula is
A=p (1+r/k)^kt
A future value
P present value
R interest rate
K compounding periods
T time
When the compounding periods are daily balance after one year is
A=1,870×(1+0.197÷365)^(365×1)
A=2,277.06
When the compounding periods are monthly balance after one year is
A=1,870×(1+0.165÷12)^(12×1)
A=2,202.99
So will save
2,277.06−2,202.99
=74.07....answer
Hope it helps!
Answer:
Step-by-step explanation:
First simplify:
Therefore we have:
![\sum\limits_{n=1}^{150}[-1-(n-1)]=\sum\limits_{n=1}^{150}(-n)=(-1)+(-2)+(-3)+...+(-150)\\\\-1,\ -2,\ -3,\ -4,\ ...,\ -150-\text{it's the arithmetic sequence}\\\text{with the common difference d = -1.}\\\\\text{The formula of a sum of terms of an arithmetic sequence:}\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\text{Substitute}\ n=150,\ a_1=-1,\ a_n=-150:\\\\S_{150}=\dfrac{-1+(-150)}{2}\cdot150=(-151)(75)=-11,325](https://tex.z-dn.net/?f=%5Csum%5Climits_%7Bn%3D1%7D%5E%7B150%7D%5B-1-%28n-1%29%5D%3D%5Csum%5Climits_%7Bn%3D1%7D%5E%7B150%7D%28-n%29%3D%28-1%29%2B%28-2%29%2B%28-3%29%2B...%2B%28-150%29%5C%5C%5C%5C-1%2C%5C%20-2%2C%5C%20-3%2C%5C%20-4%2C%5C%20...%2C%5C%20-150-%5Ctext%7Bit%27s%20the%20arithmetic%20sequence%7D%5C%5C%5Ctext%7Bwith%20the%20common%20difference%20d%20%3D%20-1.%7D%5C%5C%5C%5C%5Ctext%7BThe%20formula%20of%20a%20sum%20of%20terms%20of%20an%20arithmetic%20sequence%3A%7D%5C%5C%5C%5CS_n%3D%5Cdfrac%7Ba_1%2Ba_n%7D%7B2%7D%5Ccdot%20n%5C%5C%5C%5C%5Ctext%7BSubstitute%7D%5C%20n%3D150%2C%5C%20a_1%3D-1%2C%5C%20a_n%3D-150%3A%5C%5C%5C%5CS_%7B150%7D%3D%5Cdfrac%7B-1%2B%28-150%29%7D%7B2%7D%5Ccdot150%3D%28-151%29%2875%29%3D-11%2C325)
To solve for a variable means to ISOLATE the variable on one side of the equation. To do so, divide both sides by the coefficient of y.
Look:
2y/2 = (-x+7)/2
Answer: y = (-x + 7)/2
The answer can also be written as
y = -(x/2) + (7/2)
Both answers mean the same thing.
