Answer:
ex:Speed (S) is the ratio of the distance (D) covered to the time (t) taken.
That is, S = D/t
Suppose Andrew ran a distance D1 in 1 hour (3600 seconds) at a Speed, say S1, we have
S1 = D1/t
We can then say he ran a distance
D1 = t × S1
= 3600S1
Similarly, let's say Karleigh ran a distance
D2 = t × S2
= 3600S2
Let us compare these two, you will notice that the bigger number between S1 and S2 is going to determine the bigger number between D1 and D2.
Let's choose random numbers for S1 and S2 for clarity, say S1 = 5, S2 = 10
D1 = 3600 × 5
= 18000
D2 = 3600 × 10
= 36000
This makes D2 bigger than D1. this is an example i found on the internet.
Step-by-
hope this helps, good luck
(12x^3-4x^2+8x)/(-2x) The greatest common factor of the numerator and denominator is -2x so upon division by GCF of both numerator and denominator you have:
-6x^2+2x-4 and we can now also factor out -2
-2(3x^2-x+2)
Answer:
Logarithmic
Step-by-step explanation:
For this exmaple, x is the one increasing by a third power for every y
Logarithmic is the opposite of exponential
Exponential is when y is the one increasing by a power for every x.
Therefore this one is logarithmic.
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
