Answer:
GH¢2082.12
Step-by-step explanation:
Let "a" represent the amount invested at 12%. Then (a+580) is the amount invested at 14%. The total amount invested (t) is ...
t = (a) +(a +580) = 2a+580
Solving for a, we get
a = (t -580)/2
__
The accumulated amount from the investment at 12% is 1.12a. And the accumulated amount from the investment at 14% is 1.14(a+580). Together, these accumulated amounts total GH¢2358.60.
1.12(t -580)/2 +1.14((t -580/2 +580) = 2358.60
0.56t -0.56(580) +0.57t -0.57(580) +1.14(580) = 2358.60 . . . remove parens
1.13t + 5.8 = 2358.60 . . . . . . . . . simplify
1.13t = 2352.80 . . . . . . . . . . . . . . subtract 5.8
t = 2352.80/1.13 = 2082.12 . . . . divide by the coefficient of t
Mr. Azu's total investment was GH¢2082.12.
Answer:
C)
<h3>
log(117.50 / (117.50 - 2050(0.012) ) / log(1+0.012 ) </h3>
Step-by-step explanation:
Formula to calculate compounded monthly payments
A = R( (1-(1+r)^-n) / r)
where
r = 0.14/12
= 0.012
A = 2050
R = 117.50
n =no. of payments
2050 = 117.50 (1 - (1 + 0.012)^-n / 0.012)
cross multiplication
2050 (0.012) / 117.50 = 1 - (1 + 0.012)^-n
1 on other side
(2050 (0.012) / 117.50) - 1 = - (1+0.012)^-n
eliminating minus sign
1 - (2050 (0.012) / 117.50) = (1+0.012)^-n
LCM
(117.50 - 2050(0.012) ) / 117.50 = (1 + 0.012)^-n
power in negative
(117.50 - 2050(0.012) ) / 117.50 = 1 / (1+0.012)^n
reciprocal
117.50 / (117.50 - 2050(0.012) ) = (1+0.012)^n
taking log
log(117.50 / (117.50 - 2050(0.012) ) = log(1+0.012)^n
Answer
log(117.50 / (117.50 - 2050(0.012) ) = n log(1+0.0120)
<h3>
log(117.50 / (117.50 - 2050(0.012) ) / log(1+0.012 ) = n</h3>
About 31%
You add all of the numbers up to find out how many cans there are overall. Then take 80(cans of tuna) / overall cans and get around 31%.
You answer is 229 multiply first then add (PEMDAS)
Answer:
The diagonals bisect each other
Step-by-step explanation:
2s = 10 s+5 = 10 FJ = JH
t+12 = 18 3t = 18 EJ = JG