What is the slope of the graph?

Landon is a maintenance worker at an amusement park. He stands atop a raised platform of the park's most popular

Helga [31]

Answer:

the equation fits the situation, but the table doesn’t.

Step-by-step explanation:

For both functions, the lead roller coaster car starts at an initial height of 32 feet. This is the point (0,32), directly above Landon.

The car races past Landon’s field of vision (y = 0) after it moves a horizontal distance of 4 feet and then 8 feet with respect to the starting point. So, the correct function will have x-intercepts of 4 and 8.

Begin by checking the equation. Rewrite the equation in factored form:

f(x) = $x^{3}$ − 11 $x^{2}$ + 20x + 32

= (x + 1)( $x^{2}$ − 12x + 32)

= (x + 1)(x − 4)(x − 8)

The equation has x-intercepts of 4 and 8.

But the table only has one x-intercept, located at (4,0). The point (8,160) is well out of Landon’s line of sight.

So, the equation fits the situation, but the table doesn’t.

3 0

3 years ago

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FREE POINTS AND MYBE A BRAINLIEST JUST ONE QUESTION ARE YOU TIRED OF ASKING SONE QUESTIONS NO ONE ANSWERS AND THEN YOU CAN'T GET

natka813 [3]

Answer:

Yes totally friend, I totally hear you on that!! ^-^ Good luck for your weekend!!!

Step-by-step explanation:

3 0

3 years ago

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HELP PLEASE PLEASE!!!!!!!!

iren2701 [21]

Answer:

The measure of the third angle is equal to 34°

Step-by-step explanation:

Given that,

The measure of angle 1 is 50°.

The measure of angle 2 is 96°

We need to find the measure of the third angle. We know that the sum of angles of a triangle is equal to 180°. So,

x+50+96=180

x+146=180

x = 180-146

x = 34°

So, the measure of the third angle is equal to 34°.

8 0

3 years ago

Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen

k0ka [10]

Let $a,b,c$ be the randomly selected lengths. Without loss of generality, suppose $a[tex]P(A + B \ge C) = P(A + B - C \ge 0)$

where $A,B,C$ are independent random variables with the same uniform distribution on [0, 1].

By their mutual independence, we have

$P(A=a,B=b,C=c) = P(A=a) \times P(B=b) \times P(C=c)$

so that the joint density function is

$P(A=a,B=b,C=c) = \begin{cases}1 & \text{if }(a,b,c)\in[0,1]^3 \\ 0 & \text{otherwise}\end{cases}$

where $[0,1]^3=[0,1]\times[0,1]\times[0,1]$ is the cube with vertices at (0, 0, 0) and (1, 1, 1).

Consider the plane

$a + b - c = 0$

with $(a,b,c)\in\Bbb R^3$ . This plane passes through (0, 0, 0), (1, 0, 1), and (0, 1, 1), and thus splits up the cube into one tetrahedral region above the plane and the rest of the cube under it. (see attached plot)

The point (0, 0, 1) (the vertex of the cube above the plane) does not belong the region $a+b-c\ge0$ , since $0+0-1=-1$ . So the probability we want is the volume of the bottom "half" of the cube. We could integrate the joint density over this set, but integrating over the complement is simpler since it's a tetrahedron.

Then we have

$\displaystyle P(A+B-C\ge0) = 1 - P(A+B-C < 0) \\\\ ~~~~~~~~ = 1 - \int_0^1\int_0^{1-a}\int_{a+b}^1 P(A=a,B=b,C=c) \, dc\,db\,da \\\\ ~~~~~~~~ = 1 - \int_0^1 \int_0^{1-a} (1 - a - b) \, db \, da \\\\ ~~~~~~~~ = 1 - \int_0^1 \frac{(1-a)^2}2\,da \\\\ ~~~~~~~~ = 1 - \frac16 = \boxed{\frac56}$