Lions were introduced in the park in 2010. In 2012, the park has a population of 32 lions. A year later, in 2013, the park has 128 lions. Write an exponential function that models this situation. Let x represent the number of years since 2010 and let f(x)represent the number of lions.

You can use the fact that the given figures are expressible as 2 raised to some powers.

The exponential function modeling the number of lions to the years passed since 2010 is given as

Those functions which are expressed as some constant raised to the power of the argument called exponential function.

Since 2012  - 2010 = 2 and the number of lions = 32 =

and 2013 -2010 = 3 and the number of lions =128 =

Thus, there is increment of 2 exponent as 1 year was changed.

Thus, we have:

where a might be some constant( i took this form of exponent just by trial and error. It was already visible that one year was increasing double output, thus 2x and since there was just double increase and no other effect, there might be some non variable thing(a constant) so i thought to try some variable and see if this can work)

Since at x = 2, we get 2 raised to the power 5

thus.

Thus,

The exponential function modeling the number of lions to the years passed since 2010 is given as

You can see that this will give correct output for x = 2 and x = 3

Learn more about exponential functions here:

link

y = 2(4^(x))

Step-by-step explanation:

In 2012, the park has a population of 32 lions. A year later, in 2013, the park has 128 lions.

This means in 2012, it was 2^(5) lions while after 1 year it was 2^(7) lions

Thus, it increases every year by a ratio of 2^(2)

Thus, in 2011, it was 2^(5) - 2^(2) = 2^(3) lions

In 2010, it was 2^(3) - 2^(2) = 2

Since Lions were introduced in the park in 2010, using f(x), it means;

at x = 0, f(x) = 2

at x = 1, f(x) = 2^(3)

at x = 2, f(x) = 2^(5)

at x = 3, f(x) = 2^(7)

To find an exponential function that models the situation, let's use the generic formula: y = ab^(x)

At x = 0 and y = 2, we have;

2 = a × b^(0)

Thus,a = 2

At x = 1, y = 2^(3)

Thus; 2^(3) = 2 × b^(1)

b = 8/2

b = 4

So the exponential equation is;

y = 2(4^(x))

answer:

total cost for tiles and paints is $924.  

step-by-step explanation:

we have been given that a community hall is in the shape of a cuboid. the hall is 40m long 15m high and 3m wide.

the paint will be required for 4 walls and ceiling.

let us find area of walls and ceiling.

therefore, the area of walls and ceiling is 1410 square meters.

given: cost for 10 litre of paint is $10 and 10 litre paint covers 25 square meter. therefore,

therefore, the total painting cost is $564.

tiles will be required for floor. let us find the area of floor.

given: 1m squared floor tiles costs $3. so,

therefore, the total tiles cost is $360.

now let us find combined total cost of tiles and paint.

therefore, the combined total cost of tiles and paint is $924.