All categories

3 years ago

Find the midpoint of the line segment whose endpoints are (-2, 5) and (6, -9). (4, -4) (2, -2) (1, -4)

Mathematics

1 answer:

Rus_ich [418]3 years ago

8 0

Answer:

(2,-2)

Step-by-step explanation:

The midpoint of the line segment with endpoints $(x_1,y_1)$ and $(x_2,y_2)$ has coordinates

$\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

If the endpoints of the line segment are (-2,5) and (6,-9), then the midpoint has the coordinates

$\left(\dfrac{-2+6}{2},\dfrac{5+(-9)}{2}\right)\\ \\\left(\dfrac{4}{2},\dfrac{-4}{2}\right)\\ \\(2,-2)$

You might be interested in

Find the LCM of n^3 t^2 and nt^4. A) n 4t^6 B) n 3t^6 C) n 3t^4 D) nt^2

patriot [66]

Answer:

The correct option is C.

Step-by-step explanation:

The least common multiple (LCM) of any two numbers is the smallest number that they both divide evenly into.

The given terms are $n^3t^2$ and $nt^4$ .

The factored form of each term is

$n^3t^2=n\times n\times n\times t\times t$

$nt^4=n\times t\times t\times t\times t$

To find the LCM of given numbers, multiply all factors of both terms and common factors of both terms are multiplied once.

$LCM(n^3t^2,nt^4)=n\times n\times n\times t\times t\times t\times t$

$LCM(n^3t^2,nt^4)=n^3t^4$

The LCM of given terms is $n^3t^4$ . Therefore the correct option is C.

8 0

3 years ago

Read 2 more answers

What 15 by 15 I need to know!!

Colt1911 [192]

15 times 15 = 225. :) since i helped you, can you help me? My question: brainly.com/question/2967131

7 0

4 years ago

Read 2 more answers

120 books increased by 55%

mash [69]

55% of 120 is just 66
add that with 120, 186

so... 186 books

8 0

4 years ago

Which set of ordered pairs below is a function?

Paha777 [63]

The answer is A.

Ordered pairs that equate to a function would not have two separate points with the same x-value. In other words, every x-value must be associated with only one y-value.

5 0

3 years ago

Find the exact value of each trigonometric function for the given angle θ.

Kay [80]

Answer:

$\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=\dfrac{2}{\sqrt{3}}.$

Step-by-step explanation:

The given angle is 240 degrees.

We need to find the exact value of each trigonometric function for the given angle θ.

Since $\theta=240$ , it means θ lies in 3rd quadrant. In 3d quadrant only tan and cot are positive.

$\sin (240^\circ)=\sin (180^\circ+60^\circ)=-\sin (60^\circ)=-\dfrac{\sqrt{3}}{2}$

$\cos (240^\circ)=\cos (180^\circ+60^\circ)=-\cos (60^\circ)=-\dfrac{1}{2}$

$\tan (240^\circ)=\tan (180^\circ+60^\circ)=\tan (60^\circ)=\sqrt{3}$

$\cot (240^\circ)=\cot (180^\circ+60^\circ)=\cot (60^\circ)=\dfrac{1}{\sqrt{3}}$

$\sec (240^\circ)=\sec (180^\circ+60^\circ)=-\sec (60^\circ)=-2$

$\csc (240^\circ)=\csc (180^\circ+60^\circ)=-\csc (60^\circ)=-\dfrac{2}{\sqrt{3}}$

Therefore, $\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=-\dfrac{2}{\sqrt{3}}.$

8 0

3 years ago