A=p (1+I/k)^kn
A future value 160000
p present value ?
I interest rate 6/100=0.06
K compounded semiannual 2
T time 18-7=11years
160000=p (1+0.06/2)^(2×11)
Divide each side by (1+0.06/2)^(2×11)
P=160,000÷(1+0.06÷2)^(2×11)
P=83,502.8
Hope it helps:-)
Answer: 6 - 10 blinks would probably be a reasonable target.
Step-by-step explanation:
For point (1, -2): -2 = -1 - 1 = -2. Therefore, point (1, -2) lies on the graph of the equation.
The graph of the equation is the set of points that are solutions to the equation.
A coordinate pair on the graph of the equation is a solution to the equation.
For the point (0, -1): -1 = -(0) - 1 = 0 - 1 = -1.
Therefore, The point ( 0, −1 ) lies on the graph of the equation.
Answer:
3156
Step-by-step explanation:
- <em>Used formula:</em>
- <em>(1² + 2² + 3² + ... + n²) =1/6*n(n + 1)(2n + 1)</em>
--------
- 10²+12²+14²+......+26² =
- (2*5)²+(2*6)² + (2*7)² + ... + (2*13)² =
- 4*(5²+6²+7²+...+13²) =
- 4*(1²+2²+...+13² - (1²+2²+3²+4²)) =
- 4*(1/6*13(13+1)(2*13+1) - (1+4+9+16)) =
- 4*(1/6*13*14*27- 30) =
- 4*(819 - 30) =
- 4*789 =
- 3156
Answer:
whats the problem
Step-by-step explanation:
