Answer:
<h2><u>[D] 13.9</u></h2>
Explanation:
- <em>Pythagorean theorem: a² + b² = c²</em>
- <em>Solve for hypotenuse (side x) using: c = √a² + b²</em>
12.8² + 5.3² = 191.93
√191.93
= 13.8538803229
<em>Round the answer</em>
13.9
Answer:
v = 1/(1+i)
PV(T) = x(v + v^2 + ... + v^n) = x(1 - v^n)/i = 493
PV(G) = 3x[v + v^2 + ... + v^(2n)] = 3x[1 - v^(2n)]/i = 2748
PV(G)/PV(T) = 2748/493
{3x[1 - v^(2n)]/i}/{x(1 - v^n)/i} = 2748/493
3[1-v^(2n)]/(1-v^n) = 2748/493
Since v^(2n) = (v^n)^2 then 1 - v^(2n) = (1 - v^n)(1 + v^n)
3(1 + v^n) = 2748/493
1 + v^n = 2748/1479
v^n = 1269/1479 ~ 0.858
Step-by-step explanation:
Answer:
2 7/16 inches longer
Step-by-step explanation:
15 3/16 - 12 3/4= 243/16 - 51/4 = 243/16 - 204/16= 39/16 inches longer
39/16=2 7/16 inches longer
Hi Brielle the correct answer to your question is 3.12 x 4.0 = 12.48
Current amount in account
P=36948.61
Future value of this amount after n years at i=11% annual interest
F1=P(1+i)^n
=36948.61(1.11)^n
Future value of $3000 annual deposits after n years at i=11%
F2=A((1+i)^n-1)/i
=3000(1.11^n-1)/0.11
We'd like to have F1+F2=280000, so forming following equation:
F1+F2=280000
=>
36948.61(1.11)^n+3000(1.11^n-1)/0.11=280000
We can solve this by trial and error.
The rule of 72 tells us that money at 11% deposited will double in 72/11=6.5 years, approximately.
The initial amount of 36948.61 will become 4 times as much in 13 years, equal to approximately 147800 by then.
Meanwhile the 3000 a year for 13 years has a total of 39000. It will only grow about half as fast, namely doubling in about 13 years, or worth 78000.
Future value at 13 years = 147800+78000=225800.
That will take approximately 2 more years, or 225800*1.11^2=278000.
So our first guess is 15 years, and calculate the target amount
=36948.61(1.11)^15+3000(1.11^15-1)/0.11
=280000.01, right on.
So it takes 15.00 years to reach the goal of 280000 years.
