Answer:
![\large\boxed{4\sqrt[3]{64}=16}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B4%5Csqrt%5B3%5D%7B64%7D%3D16%7D)
Step-by-step explanation:
![\sqrt[3]{a}=b\iff b^3=a\\\\4\sqrt[3]{64}=(4)(4)=16\\\\\sqrt[3]{64}=4\ \text{because}\ 4^3=64](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Ba%7D%3Db%5Ciff%20b%5E3%3Da%5C%5C%5C%5C4%5Csqrt%5B3%5D%7B64%7D%3D%284%29%284%29%3D16%5C%5C%5C%5C%5Csqrt%5B3%5D%7B64%7D%3D4%5C%20%5Ctext%7Bbecause%7D%5C%204%5E3%3D64)
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
Answer:
The percent increase is 120%.
Answer:
finish your exam please i am a supervisor
Step-by-step explanation:
The correct answers to the question above includes:
a=.1
b=20.775
c=0.02
