Answer:
$ 5674.076
Step-by-step explanation:
The question is on compound interest
The formulae = A= P(1+ r/n) ^nt .......where P is the principal amount, r is the rate of interest in decimal, n is number of compoundings per year and t is the total number of years.
Given; P= $4,000.00 , r=12/100=0.12, n=2 and t=3
Substituting values in the equation A= P(1+ r/n) ^nt
A= 4000 ( 1+0.12/2)^2×3
A=4000(1.06)^6
A=$ 5674.08
Answer:
Roy is 10 years old at present and Joan is 5 years old at present
Step-by-step explanation:
Let
x----> Roy's age
y----> Joan's age
we know that
x=2y ----> equation A
(x+3)+(y+3)=21 ----> equation B
substitute equation A in equation B
(2y+3)+(y+3)=21
solve for y
3y+6=21
3y=21-6
3y=15
y=5 years
Find the value of x
x=2y ----> x=2(5)=10 years
therefore
Roy is 10 years old at present
Joan is 5 years old at present
Answer:
the graph shift down 135 units
Step-by-step explanation:
when there is no fixing cost:
f(x)=12x-1400
when there is fixing cost (0ne time): 12x-(1400+135)
the graph shift down 135 units
Answer:
C.
Step-by-step explanation:
A. 39 = 3 × 13; not prime
B. 33 = 3 × 11; not prime
C. 43 = 1 × 43; prime
D. 51 = 3 × 17; not prime
Answer: C.
Answer:
Below in bold.
Step-by-step explanation:
x^2 - y^2 = 11
2x^2 + y^2 = 97
From the first equation:
y^2 = x^2 - 11
Substituting in the second equation:
2x^2 + x^2 - 11 = 97
3x^2 = 108
x^2 = 36
x = 6, -6.
Substituting for x in the first equation:
(6)^2 - y^2 = 11
y^2 = 36 - 11 = 25
y = 5, -5.