After t sec. a ball tossed in the air from ground level reaches a height of h feet given by the equation: h=144t-16t^2a. what is the height of the ball after 3 sec.? 288 ft.

b. what is the maximum height the ball will reach

c. find the number of seconds the ball is in the air when it reaches 224 ft in height

d. after how many sec. will the ball hit the ground before rebound?



I am not sure which equation to use and why for the b,c,and d. If I find the ft/sec. that the ball travels (96), and set the quadratic equation to -16t^2+144t+288 or 16t^2-144t-288=0, is this correct? Or should I use the quad formula and why.

Answer:

First question:

The graph of   has a vertical asymptote at x =  and a horizontal asymptote at y =  

Second question:

The graph of equation has a horizontal asymptote at y = -3 ⇒ C

Step-by-step explanation:

The vertical asymptotes will occur at the values of x for which make the  denominator is equal to zero

The horizontal asymptotes will occur if:

• Both polynomials are the same degree, divide the coefficients of the highest degree terms
• The polynomial in the numerator is a lower degree than the denominator, the x-axis (y = 0) is the horizontal asymptote

First question:

∵

- To find the vertical asymptote equate the denominator by 0

   to find the value of x

∵ The denominator is 2 - 3x

∴ 2 - 3x = 0

- Add 3x to both sides

∴ 2 = 3x

- Divide both sides by 3

∴ = x

∴ The graph has a vertical asymptote at x =  

To find the horizontal asymptote look at the highest degree of x in both numerator and denominator

∵ The denominator and the numerator has the same degree of x

- Divide the coefficient of x of the numerator and denominator

∵ The coefficient of x in the numerator is -2

∵ The coefficient of x in the denominator is -3

∵ -2 ÷ -3 =

∴ The graph has a horizontal asymptote at y =  

The graph of   has a vertical asymptote at x =   and a horizontal asymptote at y =  

Second question:

The graph has a horizontal asymptote at y = -3

means the numerator and the denominator has same highest degree and the coefficient of the highest degree in the numerator divided by the coefficient of the highest degree in the denominator equal to -3

• In all answers the numerator and the denominator have the same highest degree

• Lets look for the coefficients of x up and down to find which one gives quotient of -3

∵ In answer A the quotient is 1 because x up and down have

  coefficient 1

∵ In answer B the quotient is because the coefficient of x

   up is 1 and down is -3

∵ In answer D the quotient is -1 because the coefficient of x

   up is 3 and down is -3

∵ In answer C the quotient is -3 because the coefficient of x up

   is -3 and down is 1

∴ The graph of equation has a horizontal asymptote at y = -3

Step-by-step explanation:

To make the function f(x) = {sin(1/x), x ≠ 0; k, x = 0} continuous at x = 0, we need to find the value of k that ensures the limit of f(x) as x approaches 0 exists and is equal to k.

First, let's find the limit of sin(1/x) as x approaches 0:

lim(x -> 0) sin(1/x)

This limit does not exist because sin(1/x) oscillates wildly as x gets closer to 0. Therefore, in order for the function to be continuous at x = 0, we need to choose k such that it compensates for the oscillations of sin(1/x) as x approaches 0.

A suitable choice for k is 0 because the limit of sin(1/x) as x approaches 0 is undefined, and setting k = 0 ensures that f(x) becomes a continuous function at x = 0.

So, the correct choice is:

d. None (k = 0)

Final answer:

The value of k that would make the function f(x) = sin(1/x) when x ≠0 and f(x) = k when x=0 continuous at x=0 doesn't exist. This is because the limit of sin(1/x) as x approaches 0 is undefined, hence the function cannot be made continuous at x = 0 for any value of k.

Explanation:

To find the value of k that makes the function continuous at x=0, we can apply the definition of continuity, which states that a function, f(x), is continuous at a certain point, x0, if three conditions are met:

• the function is defined at x0
• the limit as x approaches x0 of f(x) exists
• the limit as x approaches x0 of f(x) is equal to f(x0)

In the case of the function f(x) = sin(1/x), the value for x = 0 is undefined, but we've been given that f(0) = k. To make the function continuous at x = 0, the value of k should ideally be equal to the limit of sin(1/x) as x approaches 0.

However, as x approaches 0, sin(1/x) oscillates between -1 and 1, making the limit non-existent. Because the limit does not exist, the function is not continuous at x=0 no matter the chosen value of k. Therefore, the correct answer is (d) None.

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Answer:

lolololololololoolololo

Step-by-step explanation:

Answer:

25

Step-by-step explanation: