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3 years ago

What are the coordinates of the midpoint of the segment with endpoints at (-3, -4) and (5, 8)?

Mathematics

1 answer:

lord [1]3 years ago

8 0

Answer:

(4, ``6)

Step-by-step explanation:

The midpoint formula is

((x sub 2 - x sub 1)/2, (y sub 2 -y sub 1)/2). As shown, its written like a coordinate

Your x sub 1 is the x-coordinate of your first coordinate pair, and the rest follow the same pattern.

So, (5-(-3)/2, 8-(-4)/2)

5-(-3)=5+3=8

8-(-4)=8+4=12

(8/2, 12/2)

(4,6)

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2 years ago

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Irina-Kira [14]

Answer:

option A

Step-by-step explanation:

$\frac{x+ 3}{12x} \ \cdot \ \frac{4x}{x^2 +x - 6}\\\\= \frac{x+ 3}{12x} \ \cdot\ \frac{4x}{x^2 +3x - 2x - 6}\\\\= \frac{x+ 3}{12x} \cdot \frac{4x}{x(x + 3) -2(x+3)}\\\\= \frac{x+ 3}{12x} \cdot \frac{4x}{(x + 3)(x -2)}\\\\= \frac{1}{12x} \ \cdot \ \frac{4x}{(x - 2)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\\\= \frac{1}{3} \ \cdot \ \frac{1}{(x - 2)}\\\\= \frac{1}{3(x - 2) }$

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3 years ago

<img src="https://tex.z-dn.net/?f=prove%20that%5C%20%20%5Ctextless%20%5C%20br%20%2F%5C%20%20%5Ctextgreater%20%5C%20%5Cfrac%20%7B

inysia [295]

$\large \bigstar \frak{ } \large\underline{\sf{Solution-}}$

Consider, LHS

$\begin{gathered}\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }} \\ \end{gathered} \\ \\ \text{So, using this identity, we get} \\ \\ \begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - ( {sec}^{2}\theta - {tan}^{2}\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: \: }} \\ \end{gathered} \\$

So, using this identity, we get

$\begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - (sec\theta + tan\theta )(sec\theta - tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

can be rewritten as

$\begin{gathered}\rm\:=\:\dfrac {(\sec \theta + tan\theta ) - (sec\theta + tan\theta )(sec\theta -tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac {(\sec \theta + tan\theta ) \: \cancel{(1 - sec\theta + tan\theta )}} { \cancel{ \tan \theta - \sec \theta + 1} } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:sec\theta + tan\theta \\\end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1}{cos\theta } + \dfrac{sin\theta }{cos\theta } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1 + sin\theta }{cos\theta } \\ \end{gathered}$

<h2>Hence,</h2>

$\begin{gathered} \\ \rm\implies \:\boxed{\sf{ \:\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \:\dfrac{1 + sin\theta }{cos\theta } \: \: }} \\ \\ \end{gathered}$

$\rule{190pt}{2pt}$

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2 years ago

Multiply<br><br> 5( 4n -5)<br> tell me how

rewona [7]

Distribute

5 × 4n + 5 × -5

Simplify 5 × 4n to 20n

20n + 5 × -5

Simplify 5 × -5 to -25

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3 years ago