Question

Select "Growth" or "Decay" to classify each function.

Function Growth Decay

y=200(0.5)^2t

y=1/2(2.5)^t/6

y=(0.65)^t/4


Which equation can be used to find the solution of (1/3)^d−5 = 81 ?


A. −d−5=4


B. d + 5 = 4


C. d−5=4


D. −d+5=4


The initial number of views for a blog was 20. The number of views is growing exponentially at a rate of 20% per week.


What is the number of views expected to be four weeks from now?


Round to the nearest whole number.




Enter your answer in the box.


_____________


Which equations model exponential decay?


Select each correct answer.


A. y=0.55(0.91)^x


B. y=4.2(1.25)^x ​


C.​ ​ y=0.25(2)^x ​ ​


C.​ y=2(0.20)^x ​

Answer 1

A) y=1/2(2.5)^t/6 is the only growth function, the other two are for decay as the multipliers in the brackets are less than 1, thus indicating decay
C) 20(1.2)^4= 41 views
D) A and C as the multipliers in the brackets are less than 1, thus indicating decay

Answer 2

1) An exponential function,

`y=ab^x` is called

Growth function : If b > 1

Decay function : if 0 < b < 1

Thus, the Growth function :

`y=(1)/(2)(2.5)^(1)/(6)`

And, decay functions :

`y=200(0.5)^(2t)`
`y=(0.65)^(t)/(4)`

2) Given equation,

`((1)/(3))^(d-5)=81`

`((1)/(3))^(d-5)=(3)^4`

`((1)/(3))^(d-5)=((1)/(3))^(-4)`

`\implies d-5 = -4`

`\implies -d+5=4`

Thus, Option 'D' is correct.

3) Given,

The initial number for blog, P = 20,

Rate per week, r = 20% = 0.2

So, the number of blocks after x weeks,

`A=P(1+r)^x`

`=20(1+0.2)^x`

`=20(1.2)^x`

Hence, the number of blocks after 4 weeks,

`A=20(1.2)^4=41.472\approx 41`

4) ∵ 0.55 < 1 and 0.20 < 1

so, the exponential decay functions,

A.
`y=0.55(0.91)^x`

C.
`y=2(0.20)^x`