A quantity with an initial value of 7600 decays exponentially at a rate of 55% every 7 days. What is the value of the quantity after 2 weeks, to the nearest hundredth?

1539

Step-by-step explanation:

We solve the above question using the Exponential decay formula

= A(t) = Ao(1 - r) ^t

Ao = Initial Amount invested = 7600

r = Decay rate = 55% = 0.54

t = time in weeks = 2

Hence:

A(t) = 7600(1 - 0.55)²

A(t) = 7600 × (0.45)²

A(t) = 1539

Therefore, the value of the quantity after 2 weeks is 1539

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !

answer: c

because if you look at the chart you can see a pattern and compare it to the question he simplify out your

but i think the formula is f(x)=6x