Answer:
b(b/a)^2
Step-by-step explanation:
Given that the value of the car depreciates such that its value at the end of each year is p % less than its value at the end of the previous year and that car was worth a dollars on December 31, 2010 and was worth b dollars on December 31, 2011, then
b = a - (p% × a) = a(1-p%)
b/a = 1 - p%
p% = 1 - b/a = (a-b)/a
Let the worth of the car on December 31, 2012 be c
then
c = b - (b × p%) = b(1-p%)
Let the worth of the car on December 31, 2013 be d
then
d = c - (c × p%)
d = c(1-p%)
d = b(1-p%)(1-p%)
d = b(1-p%)^2
d = b(1- (a-b)/a)^2
d = b((a-a+b)/a)^2
d = b(b/a)^2 = b^3/a^2
The car's worth on December 31, 2013 = b(b/a)^2 = b^3/a^2
if you don't have a Unit Circle, this is a good time to get one, you'd need it, you can find many online.
Check the picture below.
Answer:
Option D. 
Step-by-step explanation:
In this problem
we have
where
B is the Final Investment Value or a Balance
P is the Principal amount of money to be invested
r is the rate of interest in decimal
t is Number of Time Periods
in this problem we have
substitute in the formula above
C can be found using pythagoras theorem. c2=a2+b2. Now, b is not given, but we know that cos(theta)=b/c=>b=c*cos(theta). Substituting b in the above relation, c2=a2+c2(cos(theta))^2=>c2=a2/(1-cos((theta))^2). c is the squareroot of c2. Hence c=sqrt(2/(1-cos((theta))^2))
Answer:
y=5x-7
Step-by-step explanation:
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