What would be the speed of Ashley's car if she traveled 140 miles in 2 hours assume that a steady speed was achieved

Which function has the following characteristics?

iren [92.7K]

Using the asymptote concept, the function with a vertical asymptote at x = 3 and an horizontal asymptote at $y = -\frac{1}{2}$ is given by:

$f(x) = -\frac{x}{2(x - 3)}$

<h3>What are the asymptotes of a function f(x)?</h3>

The vertical asymptotes are the values of x which are outside the domain, which in a fraction are the zeroes of the denominator.

The horizontal asymptote is the value of f(x) as x goes to infinity, as long as this value is different of infinity.

The vertical asymptote at x = 3 means that x = 3 is a root of the denominator, hence:

$f(x) = \frac{g(x)}{x - 3}$

The horizontal asymptote at y = -1/2 means that:

$\lim_{x \rightarrow \infty} f(x) = -\frac{1}{2}$

Which happens if $g(x) = -\frac{x}{2}$ , hence the function is:

$f(x) = -\frac{x}{2(x - 3)}$

More can be learned about asymptotes at brainly.com/question/16948935

#SPJ1

4 0

2 years ago

Given the function g(x) = 4(3)x, Section A is from x = 1 to x = 2 and Section B is from x = 3 to x = 4.

Rudik [331]

The question is defective, or at least is trying to lead you down the primrose path.
The function is linear, so the rate of change is the same no matter what interval (section) of it you're looking at.
The "rate of change" is just the slope of the function in the section. That's
(change in f(x) ) / (change in 'x') between the ends of the section.
In Section A:Length of the section = (1 - 0) = 1f(1) = 5f(0) = 0change in the value of the function = (5 - 0) = 5Rate of change = (change in the value of the function) / (size of the section) = 5/1 = 5
In Section B:Length of the section = (3 - 2) = 1 f(3) = 15f(2) = 10change in the value of the function = (15 - 10) = 5Rate of change = (change in the value of the function) / (size of the section) = 5/1 = 5
Part A:The average rate of change of each section is 5.
Part B:The average rate of change of Section B is equal to the average rate of change of Section A.
Explanation:The average rates of change in every section are equalbecause the function is linear, its graph is a straight line,and the rate of change is just the slope of the graph.

4 0

3 years ago

Train a leaves new york at 9am eastern time on monday, headed for los angeles at a constant rate of 40 miles per hour. on the sa

nlexa [21]

We can assume that B leaves Los Angeles not at noon but at 2pm Eastern Time, so 5 hours later than A leaves New York.

In 5 hours A covers 5*40=200 miles. So, A and B together should cover 3,000-200=2,800 miles.

Combined rate = 40 + 60 = 100 miles per hour.

To cover 2,800 miles they'll need 2,800/100=28 hours.

2pm Eastern Time on Monday + 28 hours = 6pm Eastern Time on Tuesday.

Answer: 6:00pm on Tuesday

7 0

3 years ago

Find number a and k so that x-2 is afactor<br>OF F(x)=x4-2ax tax-<br>X+k and F(-1)=3

Slav-nsk [51]

Question:

Find numbers a and k so that x-2 is a factor of

$f(x)=x^4-2ax^3+ax^2- x+k$

and

$f(-1)=3$

Answer:

$k = -2$ and $a=1$

Step-by-step explanation:

Given

$f(x)=x^4-2ax^3+ax^2- x+k$

$Factor:\ x - 4$

$f(-1)=3$

Required

Find a and k

For $f(-1)=3$

Substitute -1 for x

$f(x)=x^4-2ax^3+ax^2- x+k$

$f(-1) = (-1)^4 - 2a *(-1)^3 + a*(-1)^2 - (-1) + k$

$f(-1) = 1 - 2a *-1 + a*1 +1 + k$

$f(-1) = 1 +2a + a +1 + k$

$f(-1) = 2 +3a + k$

Substitute 3 for f(-1)

$3 = 2 +3a + k$

Collect Like Terms

$3 - 2 = 3a + k$

$1 = 3a + k$

Also:

If $x - 2$ is a factor, then

$f(2) = 0$

Substitute 2 for x and 0 for f(x)

$f(x)=x^4-2ax^3+ax^2- x+k$

$0 = 2^4 - 2a * 2^3 + a * 2^2 - 2 + k$

$0 = 16 - 2a * 8 + a * 4 - 2 + k$

$0 = 16 - 16a + 4a - 2 + k$

$0 = 16 - 12a - 2 + k$

Collect Like Terms

$2 - 16 = - 12a + k$

$-14 = k - 12a$

$k = 12a - 14$

Substitute 12a - 14 for k in $1 = 3a + k$

$1 = 3a + 12a - 14$

$1 = 15a - 14$

Collect Like Terms

$15a = 1 + 14$

$15a = 15$

Solve for a

$a = \frac{15}{15}$

$a=1$

Substitute 1 for a in $k = 12a - 14$

$k = 12 * 1 - 14$

$k = 12 - 14$

$k = -2$

3 0

2 years ago

If I wanted to estimate the √99, the first step would be to find the two squares that 99 lies on the numberline. I could then th

____ [38]

Answer:

Step-by-step explanation:

Smaller perfect squares near 99 is 81

Larger perfect square near 99 is 100

First step would be to find the two perfect squares that lies between on the number line. I could then think about the number 99 and how close it is to the smaller perfect square and the larger perfect square. That could tell me how far above or below the of the two perfect squares 99 lies on the number line. I could then take the square root of the perfect squares to see how I would estimate the square root of 99. The √99 is almost 10.

81 < 99 < 100

√81 < √99 < √100

8 < √99 < 10

So, √99 is almost 10.

8 0

1 year ago

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