What is the Value of:

Find the complex zeros of the following polynomial function.

GalinKa [24]

Answer:

$x=\frac{-1\pm\sqrt{3}i}{2}$

Step-by-step explanation:

$f(x)=x^3-1$

$0=x^3-1$

$0=(x^2+2x+1)(x-1)$

$x-1=0$

$x=1$ <-- Real Solution

$x^2+x+1=0$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-2\pm\sqrt{2^2-4(1)(1)}}{2(1)}$

$x=\frac{-1\pm\sqrt{1^2-4(1)(1)}}{2(1)}$

$x=\frac{-1\pm\sqrt{1-4}}{2}$

$x=\frac{-1\pm\sqrt{-3}}{2}$

$x=\frac{-1\pm\sqrt{3}i}{2}$ <-- Complex Solutions

6 0

3 years ago

Suppose that a box contains 7 cameras and that 5 of them are defective. A sample of 2 cameras is selected at random without repl

irinina [24]

Answer:

E(x) = 1.43 (Approx)

Step-by-step explanation:

Given:

Total number of camera = 7

Defective camera = 5

Sample selected = 2

Computation:

when x = 0

P(x=0) = 2/7 × 1/6 = 2/42

P(x=1) = [2/7 × 5/6] + [5/7 × 2/6] = 20/42

P(x=2) = 5/7 × 4/6 = 20/42

So,

E(x) = [0×2/42] + [1×20/42] + [2×20/42]

E(x) = 1.43 (Approx)

8 0

3 years ago

FIRST GETS BRAINIEST Write a story problem in which you would have to multiply 5 × (-1/3)

AveGali [126]

Answer:

5 students from Westville Elementary school re on a field trip. They come to a hot dog stand and no one has money. One student's parents has a tab at the hot dog stand. So the student, along with the four others, decide to order food and put it on the tab. Each hot dog is 1/3 of a dollar. They would like to know how much they are in debt.

5 times the negative (-1/3) for each hot dog.

-1 and 2/3 dollars

Glad I was able to help!!

4 0

3 years ago

Read 2 more answers

F⃗ (x,y)=−yi⃗ +xj⃗ f→(x,y)=−yi→+xj→ and cc is the line segment from point p=(5,0)p=(5,0) to q=(0,2)q=(0,2). (a) find a vector pa

DerKrebs [107]

a. Parameterize $C$ by

$\vec r(t)=(1-t)(5\,\vec\imath)+t(2\,\vec\jmath)=(5-5t)\,\vec\imath+2t\,\vec\jmath$

with $0\le t\le1$ .

b/c. The line integral of $\vec F(x,y)=-y\,\vec\imath+x\,\vec\jmath$ over $C$ is

$\displaystyle\int_C\vec F(x,y)\cdot\mathrm d\vec r=\int_0^1\vec F(x(t),y(t))\cdot\frac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt$

$=\displaystyle\int_0^1(-2t\,\vec\imath+(5-5t)\,\vec\jmath)\cdot(-5\,\vec\imath+2\,\vec\jmath)\,\mathrm dt$

$=\displaystyle\int_0^1(10t+(10-10t))\,\mathrm dt$

$=\displaystyle10\int_0^1\mathrm dt=\boxed{10}$

d. Notice that we can write the line integral as

$\displaystyle\int_C\vecF\cdot\mathrm d\vec r=\int_C(-y\,\mathrm dx+x\,\mathrm dy)$

By Green's theorem, the line integral is equivalent to

$\displaystyle\iint_D\left(\frac{\partial x}{\partial x}-\frac{\partial(-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=2\iint_D\mathrm dx\,\mathrm dy$

where $D$ is the triangle bounded by $C$ , and this integral is simply twice the area of $D$ . $D$ is a right triangle with legs 2 and 5, so its area is 5 and the integral's value is 10.

4 0

3 years ago

For two events and , the probability that occurs is 0.8, the probability that occurs is 0.4, and the probability that both occur

sergey [27]

Answer:

P(B|A)=0.25 , P(A|B) =0.5

Step-by-step explanation:

The question provides the following data:

P(A)= 0.8

P(B)= 0.4

P(A∩B) = 0.2

Since the question does not mention which of the conditional probabilities need to be found out, I will show the working to calculate both of them.

To calculate the probability that event B will occur given that A has already occurred (P(B|A) is read as the probability of event B given A) can be calculated as:

P(B|A) = P(A∩B)/P(A)

= (0.2) / (0.8)

P(B|A)=0.25

To calculate the probability that event A will occur given that B has already occurred (P(A|B) is read as the probability of event A given B) can be calculated as:

P(A|B) = P(A∩B)/P(B)

= (0.2)/(0.4)

P(A|B) =0.5

7 0

4 years ago