Which of the graphs above is the graph of the equation below? Y=x^3-6x^2+11x-6=(x-3)(x-2)(x-1)

Answers

Answer 1

Answer:

Answer:

answer Z

Step-by-step explanation:

Look for a graph that contains the following zeros: x = 1, x = 2 , x= 3, following the info derived by the binomial factors that the function contains. Also look ate the fact that the function in question has for leading term positive $x^3$ , then this function must go towards plus infinity when x becomes large. This is the case for the graph option Z (the last graph of the group)

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Which of the following are true? I. Two events are disjoint if they can’t both occur at the same time.
II. Two events are independent if they have the same probability.
III. An event and its complement have probabilities that always add up to 1.

A) I only
B) II only
C) III only
D) I and II only
E) I and III only

Answers

Option E) I and III only is the correct answer

Step-by-step explanation:

First of all we have to define the disjoint events and independent events

Disjoint events:

Two events are said to be mutually exclusive or disjoint if they cannot occur at the same time. There will be no common elements in their outcomes.

Independent Events:

Two events are said to be independent if the probability of occurrence of one event doesn't affect the probability of occurrence of other event.

Moreover,

The sum of probability of an events occurrence and its complement is 1.

So by looking at the definitions we can say that statement I and statement III are true

Hence,

Option E) I and III only is the correct answer

Keywords: Probability, events

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The figure shows eight congruent triangles made by dividing a square that has an area of 64 cm2. What is the area of ABH? A. 20 cm2

B. 16 cm2

C. 8 cm2

D. 6 cm2

E. 4 cm2

Answers

The answer is C) 8 cm squared.

That is because 64/8 = 8 cm squared. Hope this helps!!!

Answer:

C. 8 cm2

Step-by-step explanation:

If the big square is 64 cm2, then each side is 8 cm (8cm x 8cm = 64 cm2)

Each of the small squares then has both sides of half the side of the big one, then, each small square has 4cm x 4cm = 16 cm2

Ihen, each small triangle's area (such as ABH) is (b.h)/2, being the base and the height equal to the sides of the small square. Each small triangle is (4cm x 4cm)/2=8cm2

What is the vortex of the quadratic function fox (x-8)(x-2)?

Answers

Answer:

Vertex is (5,-9)

Step-by-step explanation:

Answer: The vertex is (5,-9).

Step-by-step explanation:

professor must randomly select 4 students to participate in a mock debate. There are 18 students in his class. In how many different ways can these students be selected, if the order of selection does not matter?

Answers

18 divided by 4 gives 4r2 so we could do 32 different ways

Select two polynomials that have a sum of 16n^3+13n^2+7n-10 A. 8n^3 - 5n +11n^2 - 5 B. 7n^2 + 8n - n^3 + 2 C. 15n^2 - 10n + 3n^3 - 4 D. 2n^2 + 12n - 5 + 8n^3

Answers

Option A and Option D are correct, $8n^3 - 5n +11n^2 - 5$ and $2n^2 + 12n - 5 + 8n^3$ are added to form the polynomial $16n^3 + 13n^2 + 7n - 10.$

To find two polynomials that have a sum of $16n^3 + 13n^2 + 7n - 10.$

Add the coefficients of the corresponding terms in each polynomial and see if they match the coefficients of the given polynomial.

In the given options, let us consider two polynomials:

$8n^3 - 5n +11n^2 - 5$ and $2n^2 + 12n - 5 + 8n^3$

Add these two polynomials:

$8n^3 - 5n +11n^2 - 5 + 2n^2 + 12n - 5 + 8n^3$

Group the like terms:

$8n^3+8n^3+11n^2+2n^2-5n+12n-5-5$

Combine like terms:

$16n^3 + 13n^2 + 7n - 10.$

Hence, the two polynomials that have a sum of $16n^3 + 13n^2 + 7n - 10$ are $8n^3 - 5n +11n^2 - 5$ and $2n^2 + 12n - 5 + 8n^3$ . Option A and D are correct.

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Jill is building a wooden raised bed garden for her vegetable she wants the frame to have dimension X feet by X +3 feet if she wants the garden to have a maximum area of 50 ft.² what are the possible values of X

Answers

Answer: 0ft < X ≤ 5.73ft

Step-by-step explanation:

In this case, we have a rectangle.

For a rectangle of length L and width W, the area is:

A = L*W.

And we have:

L = X

W = X + 3ft.

Then the area will be:

A = X*(X + 3ft) = X^2 + 3ft*X.

And we want to have a maximum area of 50ft^2.

Then we can write:

A = X^2 + 3ft*X ≤ 50ft^2

Now let's solve this for X.

Now, the first thing we can see is that both coefficients in our quadratic equation are positive, so as the absolute value of X increases, also does the whole equation.

Then makes sense start for the upper limit of X, this is when:

X^2 + 3ft*X = 50ft^2.

Now we can solve the quadratic equation:

X^2 + 3ft*X - 50ft^2 = 0

Applying the Bhaskara formula, the solutions are:

$X = (-3ft +- √((3ft)^2 - 4*1*(-50ft^2)) )/(2) = (-3ft +-14.46ft )/(2)$