All categories

2 years ago

Help Please!!!What is the slope of the line passing through the points (-1, 7) and (3, 4)?

Mathematics

1 answer:

kari74 [83]2 years ago

7 0

Answer:

<h3>The answer is option C</h3>

Step-by-step explanation:

The slope of a line given two points can be found by using the formula

$m = \frac{ y_2 - y _ 1}{x_ 2 - x_ 1} \\$

where

(x1 , y1) and (x2 , y2) are the points

We have

$m = \frac{4 - 7}{3 - - 1} = \frac{ - 3}{3 + 1} = - \frac{3}{4} \\$

We have the final answer as

$- \frac{3}{4} \\$

Hope this helps you

You might be interested in

Please answer number 3 I will give brainliest thank you!

vaieri [72.5K]

Answer:

Skewed to right

Step-by-step explanation:

there is no explanation, it just is, just like how 1+1 is 2

Brainleist! as you promised!

4 0

2 years ago

Read 2 more answers

Jere is thinking of a number less than 28.431 and greater than 28.404. Which of the following could be Jere's number? 28.435, 28

madam [21]

28.430, 28.435... those are your answers

7 0

2 years ago

Read 2 more answers

PLease help please please please please help help help me me em e

juin [17]

1) Which of the following is an example of a dilation?
The correct answer is A) The hearts.

2) If triangle ABC is reflected across line m, the new coordinates will be located at A'(2, -4) B'(7,-7) C'(-2, -7).
The answer is False.

I hope this helps! I can't really see #3. Sorry. But I got #'s 1 & 2 though!

4 0

2 years ago

Read 2 more answers

Consider the following function.

Kryger [21]

Answer:

See below

Step-by-step explanation:

I assume the function is $f(x)=1+\frac{5}{x}-\frac{4}{x^2}$

A) The vertical asymptotes are located where the denominator is equal to 0. Therefore, $x=0$ is the only vertical asymptote.

B) Set the first derivative equal to 0 and solve:

$f(x)=1+\frac{5}{x}-\frac{4}{x^2}$

$f'(x)=-\frac{5}{x^2}+\frac{8}{x^3}$

$0=-\frac{5}{x^2}+\frac{8}{x^3}$

$0=-5x+8$

$5x=8$

$x=\frac{8}{5}$

Now we test where the function is increasing and decreasing on each side. I will use 2 and 1 to test this:

$f'(2)=-\frac{5}{2^2}+\frac{8}{2^3}=-\frac{5}{4}+\frac{8}{8}=-\frac{5}{4}+1=-\frac{1}{4}$

$f'(1)=-\frac{5}{1^2}+\frac{8}{1^3}=-\frac{5}{1}+\frac{8}{1}=-5+8=3$

Therefore, the function increases on the interval $(0,\frac{8}{5})$ and decreases on the interval $(-\infty,0),(\frac{8}{5},\infty)$ .

C) Since we determined that the slope is 0 when $x=\frac{8}{5}$ from the first derivative, plugging it into the original function tells us where the extrema are. Therefore, $f(\frac{8}{5})=1+\frac{5}{\frac{8}{5}}-\frac{4}{\frac{8}{5}^2 }=\frac{41}{16}$ , meaning there's an extreme at the point $(\frac{8}{5},\frac{41}{16})$ , but is it a maximum or minimum? To answer that, we will plug in $x=\frac{8}{5}$ into the second derivative which is $f''(x)=\frac{10}{x^3}-\frac{24}{x^4}$ . If $f''(x)>0$ , then it's a minimum. If $f''(x)$ , then it's a maximum. If $f''(x)=0$ , the test fails. So, $f''(\frac{8}{5})=\frac{10}{\frac{8}{5}^3}-\frac{24}{\frac{8}{5}^4}=-\frac{625}{512}$ , which means $(\frac{8}{5},\frac{41}{16})$ is a local maximum.

D) Now set the second derivative equal to 0 and solve:

$f''(x)=\frac{10}{x^3}-\frac{24}{x^4}$

$0=\frac{10}{x^3}-\frac{24}{x^4}$

$0=10x-24$

$-10x=-24$

$x=\frac{24}{10}$

$x=\frac{12}{5}$

We then test where $f''(x)$ is negative or positive by plugging in test values. I will use -1 and 3 to test this:

$f''(-1)=\frac{10}{(-1)^3}-\frac{24}{(-1)^4}=-34$ , so the function is concave down on the interval $(-\infty,0)\cup(0,\frac{12}{5})$

$f''(3)=\frac{10}{3^3}-\frac{24}{3^4}=\frac{2}{27}>0$ , so the function is concave up on the interval $(\frac{12}{5},\infty)$

The inflection point is where concavity changes, which can be determined by plugging in $x=\frac{12}{5}$ into the original function, which would be $f(\frac{12}{5})=1+\frac{5}{\frac{12}{5}}+\frac{4}{\frac{12}{5}^2 }=\frac{43}{18}$ , or $(\frac{12}{5},\frac{43}{18})$ .