All categories

3 years ago

Which function represents exponential decay?

Mathematics

2 answers:

n200080 [17]3 years ago

8 0

<h2>Answer:</h2>

Option: D is the correct answer.

D) $f(x)=4\cdot (\dfrac{2}{3})^x$

<h2>Step-by-step explanation:</h2>

We know that a exponential function is in general represented by:

$f(x)=ab^x$

where a>0 and b is called the base of the function and x is the exponent.

and if b>1 then the function is a exponential growth function

and 0<b<1 then the function is a exponential decay function.

$f(x)=\dfrac{1}{2}\cdot (\dfrac{3}{2})^x$

This is a exponential growth function.

Since,

$b=\dfrac{3}{2}>1$

$f(x)=\dfrac{1}{2}\cdot (\dfrac{-3}{2})^x$

This is not a exponential function because b is not strictly greater than zero.

$f(x)=4\cdot (\dfrac{-2}{3})^x$

This is also not a exponential function because b is not strictly greater than zero.

$f(x)=4\cdot (\dfrac{2}{3})^x$

This is a exponential decay function.

Because it fulfills the condition of the exponential decay function.

Since,

$b=\dfrac{2}{3}$

Zigmanuir [339]3 years ago

7 0

Answer:

Step-by-step explanation:

Exponential decay occurs when the base is less than 1 but greater than 0.

3/2 = 1.5 and is greater than 1.

-3/2 is not greater than 0 and is not exponential

-2/3 is not greater than 0 and is not exponential

2/3 is less than 1 and greater than 0. This is decay.

You might be interested in

A monkey is swinging from a tree. On each swing, she travels along an arc that is 75% as long as the previous swing's arc. The t

mr_godi [17]

Answer:

Total length of the first swing=64 m

Step-by-step explanation:

The total length of all four swings can be expressed as;

Total length of all 4 swings=Length of first swing+length of second swing+length of third swing+length of fourth swing

where;

Total length of all 4 swings=175 m

Length of first swing=x

Length of second swing=75% of length of first swing=(75/100)×x=0.75 x

Length of third swing=75% of length of second swing

Length of third swing=(75/100)×0.75 x=0.5625 x

Length of fourth swing=75% of length of third swing

Length of fourth swing=(75/100)×0.5625 x=0.421875 x

replacing;

Total length of all 4 swings

175=x+0.75 x+0.5625 x+0.421875 x

2.734375 x=175

x=175/2.734375

x=64

Total length of the first swing=x=64 m

3 0

4 years ago

Read 2 more answers

A circle with radius of \greenD{5\,\text{cm}}5cmstart color #1fab54, 5, start text, c, m, end text, end color #1fab54 sits insid

Marat540 [252]

Answer:

The area of the shaded region is 42.50 cm².

Step-by-step explanation:

Consider the figure below.

The radius of the circle is, <em>r</em> = 5 cm.

The sides of the rectangle are:

<em>l</em> = 11 cm

<em>b</em> = 11 cm.

Compute the area of the shaded region as follows:

Area of the shaded region = Area of rectangle - Area of circle

$=[l\times b]-[\pi r^{2}]\\\\=[11\times 11]-[3.14\times 5\times 5]\\\\=121-78.50\\\\=42.50$

Thus, the area of the shaded region is 42.50 cm².

7 0

3 years ago

Read 2 more answers

Suppose that f: R --> R is a continuous function such that f(x +y) = f(x)+ f(y) for all x, yER Prove that there exists KeR su

Pachacha [2.7K]

<h2>Answer with explanation:</h2>

It is given that:

f: R → R is a continuous function such that:

$f(x+y)=f(x)+f(y)------(1)$ ∀ x,y ∈ R

Now, let us assume f(1)=k

Also,

f(0)=0

( Since,

f(0)=f(0+0)

i.e.

f(0)=f(0)+f(0)

By using property (1)

Also,

f(0)=2f(0)

i.e.

2f(0)-f(0)=0

i.e.

f(0)=0 )

Also,

f(2)=f(1+1)

i.e.

f(2)=f(1)+f(1) ( By using property (1) )

i.e.

f(2)=2f(1)

i.e.

f(2)=2k

Similarly for any m ∈ N

f(m)=f(1+1+1+...+1)

i.e.

f(m)=f(1)+f(1)+f(1)+.......+f(1) (m times)

i.e.

f(m)=mf(1)

i.e.

f(m)=mk

Now,

$f(1)=f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=f(\dfrac{1}{n})+f(\dfrac{1}{n})+....+f(\dfrac{1}{n})\\\\\\i.e.\\\\\\f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=nf(\dfrac{1}{n})=f(1)=k\\\\\\i.e.\\\\\\f(\dfrac{1}{n})=k\cdot \dfrac{1}{n}$

Also,

when x∈ Q

i.e. $x=\dfrac{p}{q}$

Then,

$f(\dfrac{p}{q})=f(\dfrac{1}{q})+f(\dfrac{1}{q})+.....+f(\dfrac{1}{q})=pf(\dfrac{1}{q})\\\\i.e.\\\\f(\dfrac{p}{q})=p\dfrac{k}{q}\\\\i.e.\\\\f(\dfrac{p}{q})=k\dfrac{p}{q}\\\\i.e.\\\\f(x)=kx\ for\ all\ x\ belongs\ to\ Q$