2 years ago

What is the length of the line segment with end points (5, -7) and (5, 11)

Mathematics

1 answer:

Law Incorporation [45]2 years ago

4 0

Answer:

Step-by-step explanation:

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

We have the points (5, -7) and (5, 11). Substitute:

$d=\sqrt{(5-5)^2+(11-(-7))^2}=\sqrt{0^2+18^2}=\sqrt{18^2}=18$

You might be interested in

Find the value of x. 600x = 360000

Kruka [31]

600x = 360000
So, the x should mean 'times'.
600 x 600 = 360000.
x = 600.

5 0

2 years ago

Read 2 more answers

PLZ HELP URGENT!!!!!!!

GrogVix [38]

Answer:

t=1

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

This is geometry. Please help explain this.

lina2011 [118]

Sorry I don’t now Gemini

5 0

2 years ago

What is 7.2km/h in m/s?

lukranit [14]

Answer: 2 meters/ seconds

Step-by-step explanation

3 0

2 years ago

Read 2 more answers

Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that

kherson [118]

Answer:

The answer is

$f(x) = {\displaystyle 8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }$

Step-by-step explanation:

Remember that Taylor says that

$f(x) = {\displaystyle \sum\limits_{k=0}^{\infty} \frac{f^{(k)}(a) }{k!}(x-a)^k }$

For this case

$f^{(0)} (-2) = 8(-2)-3(-2)^3 = 8\\f^{(1)} (-2) = 8-3(3)(-2)^2 = -28\\f^{(2)} (-2) = -3(3)2(-2) = -36\\f^{(2)} (-2) = -3(3)2 = -18$

$f(x) = {\displaystyle 8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }$

5 0

3 years ago

What is the length of the line segment with end points (5, -7) and (5, 11)​

What is the length of the line segment with end points (5, -7) and (5, 11)