Translate the following algebraic<br>equation 3g

What are the solutions to the equation?<br> 7x^3=28x

11Alexandr11 [23.1K]

7x³ = 28x is our equation. We want its solutions.

When you have x and different powers, set the whole thing equal to zero.

7x³ = 28x

7x³ - 28x = 0

Now notice there's a common x in both terms. Let's factor it out.

x (7x² - 28) = 0

As 7 is a factor of 7 and 28, it too can be factored out.

x (7) (x² - 4) = 0

We can further factor x² - 4. We want a pair of numbers that multiply to 4 and whose sum is zero. The pairs are 1 and 4, 2 and 2. If we add 2 and -2 we get zero.

x (7) (x - 2) (x + 2) = 0

Now we use the Zero Product Property - if some product multiplies to zero, so do its pieces.

x = 0 -----> so x = 0

7 = 0 -----> no solution

x - 2 = 0 ----> so x = 2 after adding 2 to both sides

x + 2 = 0 ---> so = x - 2 after subtracting 2 to both sides

Thus the solutions are x = 0, x = 2, x = -2.

6 0

2 years ago

What's the slope of the graph

kotykmax [81]

Answer:

1/-2

Step-by-step explanation:

To find the slope:

rise/run

From point to point

So....

From (-1,0)

you go up 2(2)

then left 4(-4)

to get to (-5,2)

so 2/-4 reduced is 1/-2

6 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$\\=E(X-\mu_X)^2 +E(Y- \mu_Y)^2+2E(X-\mu_X)(Y- \mu_Y)]\\=E[(X-E(X)]^2 +E[Y- E(Y)]^2+2Cov (X,Y)$