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2 years ago

2. Consider the function, f(x)=x^3+x^2-9x-9

Mathematics

1 answer:

Sedaia [141]2 years ago

5 0

The intercepts of the graph are:

x-axis interception: $\left(-1,\:0\right),\:\left(-3,\:0\right),\:\left(3,\:0\right)$ .

y-axis interception: $\left(0,\:-9\right)$ .

See the graph of the function $f(x)=x^3+x^2-9x-9$ in the attached image.

<h3>Constructing a graph</h3>

For constructing a graph we have the following steps:

Determine the range of values for x of your graph.

For this exercise, for example, we can define a range -4<x<4. In others words, the values of x will be in this interval.

Determine the points

Replace these x-values in the given equation. For example:

When x=-4, we will have: $\left(-4\right)^3+\left(-4\right)^2-9\left(-4\right)-9=-21$ . Do this for the all x-values of your ranges.

See the results for this step in the attached table.

Draw the graph

Mark the points <u>x</u> and<u> y</u> that you found in the last step. After that, connect the dots to draw the graph.

The attached image shows the graph for the given function.

<h3>Find the x- and y-intercepts</h3>

The intercepts are points that crosses the axes of your plot. From your graph is possible to see:

x-axis interception points (y=f(x)=0) are: $\left(-1,\:0\right),\:\left(-3,\:0\right),\:\left(3,\:0\right)$ .

y-axis interception point (x=0) is: $\left(0,\:-9\right)$ .

Learn more about intercepts of the graph here:

brainly.com/question/4504979

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3 years ago

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Which point is a solution to the inequality shown in this graph? (3,-1) (-3,-3)

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3 years ago

Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$

Viktor [21]

When $n=0$ , we have

$5^{3^0} + 1 = 5^1 + 1 = 6$

$3^{0 + 1} = 3^1 = 3$

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for $n=k$ , that

$3^{k + 1} \mid 5^{3^k} + 1$

Now for $n=k+1$ , we have

$5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)$

so we know the left side is at least divisible by $3^{k+1}$ by our assumption.

It remains to show that

$3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1$

which is easily done with Fermat's little theorem. It says

$a^p \equiv a \pmod p$

where $p$ is prime and $a$ is any integer. Then for any positive integer $x$ ,

$5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3$

Furthermore,

$5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3$

which goes all the way down to

$5^{3^k} \equiv 5 \pmod 3$

So, we find that

$\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3$

QED

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2 years ago

Consider two competing firms in a declining industry that cannot support both firms profitably. Each firm has three possible cho

yaroslaw [1]

Answer:

a) attached below

b) ( T,T )

c) The Pure-strategy Nash equilibria are : ( N,E ) and ( E,N )

d) The mixed-strategy Nash equilibrium for Firm 1 = ( 1/3 , 0, 2/3 )

while the mixed -strategy Nash equilibrium for Firm 2 = ( 1/3 , 0, 2/3 )

Step-by-step explanation:

A) write down the game in matrix form

let: E = exit at the industry immediately

T = exit at the end of the quarter

N = exit at the end of the next quarter

matrix is attached below

B) weakly dominated strategies is ( T,T )

C) Find the pure-strategy Nash equilibria

The Pure-strategy Nash equilibria are : ( N,E ) and ( E,N )

D ) Find the unique mixed-strategy Nash equilibrium

The mixed-strategy Nash equilibrium for Firm 1 = ( 1/3 , 0, 2/3 )

while the mixed -strategy Nash equilibrium for Firm 2 = ( 1/3 , 0, 2/3 ) since T is weakly dominated then the mixed strategy will be NE

Assume that P is the probability of firm 1 exiting immediately ( E )

and q is the probability of firm 1 staying till next term ( N ) ∴ q = 1 - P.

hence the expected utility of firm 2 choosing E = 0 while the expected utility of choosing N = 4p - 2q .

The expected utilities of E and N to firm 2 =

0 = 4p - 2q = 4p - 2 ( 1-p) = 6p -2 which means : p = 1/3 , q = 2/3

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3 years ago

A baseball team played 130 games in a season. If it won sixteen more games

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Answer:

The team has 92 wins and 38 losses.

Step-by-step explanation:

In a season 130 games are played by a baseball team.

Let us assume that the team wins x games and losses y games.

Therefore, we can write, x + y=130....... (1)

It is given that if the team won sixteen more games twice as many games as it lost.

So, we can conclude that x =2y+16 ......(2)

Hence, solving equations (1) and (2) we get, (2y+16) + y =130

⇒ 3y = 114

⇒ y =38

And from equation (2), x = 2×38 + 16 = 92

Therefore, the team has 92 wins and 38 losses. (Answer)

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3 years ago