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2 years ago

Write an equation in slope-intercept form for the line with slope 8 and y-intercept -9

Mathematics

1 answer:

sdas [7]2 years ago

7 0

Answer:y=8x-9

Step-by-step explanation:

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In the figure below, BD and EC are diameters of circle P.

Zinaida [17]

Answer: 333 degrees

Step-by-step explanation:

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3 years ago

In ABC, what is the value of x?<br><br> Pls hurry im on a time limit here

kiruha [24]

Answer:

Step-by-step explanation:

By exterior angle theorem:

$(2x + 12) \degree = 112 \degree + 180 \degree - 2x \degree \\ \\ (2x + 12) \degree = 292 \degree - 2x \degree \\ \\ (2x + 12) \degree + 2x \degree = 292 \degree \\ \\ (2x + 12 + 2x) \degree = 292 \degree \\ \\ (4x + 12) \degree = 292 \degree \\ \\ 4x + 12 = 292 \\ \\ 4x = 292 - 12 \\ \\ 4x = 280 \\ \\ x = \frac{280}{4} \\ \\ x = 70$

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2 years ago

Graph the line that represents this equation:<br> y = -5.1 +2

Mrac [35]

That's for the equation I got from the picture

4 0

3 years ago

Find the base of a parallelogram with the area of 60 in.² and height of 4 inches use the formula for the area of the parallelogr

Alexxx [7]

A = bh
60 in.² = b(4)
60/4 = b
b = 15 inches

Hope this helped!

7 0

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>

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

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

5 0

2 years ago