All categories

2 years ago

Find the rational solution of x3 – 3x2 + x + 5 = 0. Show your use of the Rational Zero Theorem to receive credit!

1 answer:

5 0

A zero of a function is a number that, when plugged in for the variable, makes the function equal to zero.
P ( x ) = 0
As P ( - 1 ) = 0:
( x³ - 3 x² + x + 5 ) : ( x + 1 ) = x² - 4 x + 5
-x³ - x²
--------------
- 4 x² + x
4 x²+ 4 x
--------------
5 x + 5
-5 x - 5
------------
0
The remaining trinomial: x² - 4 x + 5 doesn´t have rational roots.
Answer: x = -1

You might be interested in

Henry is staining the wooden floor of a court. The court is in the shape of a rectangle. Its length is 54 feet and its width is

tensa zangetsu [6.8K]

Answer:

Henry would need 12 cans to cover the court

Step-by-step explanation:

1: 30 times 54 = 1620

2: 1620 divided by 135 = 12

8 0

3 years ago

Tomas has $45.00 to purchase favors for a party. If the favors are $3.75 each, how many favors can Tomas buy?

elena-14-01-66 [18.8K]

12 is right^^^^^^^^^^^^

5 0

2 years ago

Read 2 more answers

Find f(-5) f(x) = -4x + 3 A) -17 B) -7 C) 12 D) 23

Andrew [12]

Answer:

Step-by-step explanation:

=tyhgff GJ jhvcvvbvcccvjuuuu hgfdvghg jhvgfgg bbvcfff bbvcfff bbvcfff hbbjggg hhfccgyhb 5th and 3rd of November in a rectangle has been released in December 5th of July in a square of the area and I am very interested to the area and I look for the position and am looking forward for the position of your firm in my work to be a great fit and would love ms your help and if I could help with any other things

3 0

2 years ago

Let X and Y be discrete random variables. Let E[X] and var[X] be the expected value and variance, respectively, of a random vari

Ulleksa [173]

Answer:

(a) $E[X+Y]=E[X]+E[Y]$

(b) $Var(X+Y)=Var(X)+Var(Y)$

Step-by-step explanation:

Let X and Y be discrete random variables and E(X) and Var(X) are the Expected Values and Variance of X respectively.

(a)We want to show that E[X + Y ] = E[X] + E[Y ].

When we have two random variables instead of one, we consider their joint distribution function.

For a function f(X,Y) of discrete variables X and Y, we can define

$E[f(X,Y)]=\sum_{x,y}f(x,y)\cdot P(X=x, Y=y).$

Since f(X,Y)=X+Y

$E[X+Y]=\sum_{x,y}(x+y)P(X=x,Y=y)\\=\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y).$

Let us look at the first of these sums.

$\sum_{x,y}xP(X=x,Y=y)\\=\sum_{x}x\sum_{y}P(X=x,Y=y)\\\text{Taking Marginal distribution of x}\\=\sum_{x}xP(X=x)=E[X].$

Similarly,

$\sum_{x,y}yP(X=x,Y=y)\\=\sum_{y}y\sum_{x}P(X=x,Y=y)\\\text{Taking Marginal distribution of y}\\=\sum_{y}yP(Y=y)=E[Y].$

Combining these two gives the formula:

$\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y) =E(X)+E(Y)$

Therefore:

$E[X+Y]=E[X]+E[Y] \text{ as required.}$

(b)We want to show that if X and Y are independent random variables, then:

$Var(X+Y)=Var(X)+Var(Y)$

By definition of Variance, we have that:

$Var(X+Y)=E(X+Y-E[X+Y]^2)$

$=E[(X-\mu_X +Y- \mu_Y)^2]\\=E[(X-\mu_X)^2 +(Y- \mu_Y)^2+2(X-\mu_X)(Y- \mu_Y)]\\$Since we have shown that expectation is linear$\\=E(X-\mu_X)^2 +E(Y- \mu_Y)^2+2E(X-\mu_X)(Y- \mu_Y)]\\=E[(X-E(X)]^2 +E[Y- E(Y)]^2+2Cov (X,Y)$