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Natali5045456 [20]

3 years ago

10

write the equation of the line using the given information. type your awnser in slope intercept form,y=mx+b

1 answer:

faltersainse [42]3 years ago

5 0

Answer:

y=3/-1x+4

Step-by-step explanation:

Okay so first thing you need to know is that if there is an ordered pair (x, y) where x is 0, your y-intercept is your y. For example, your problem has (0, 4) your x is 0 and your y is 4. Therefore your y-intercept is 4 which is the b. To find your mx, or slope, you need to do (y2-y1)/(x2-x1). Your y2 will be your y in your second ordered pair and your y1 will be in your y in the first ordered pair. Same for your x. So, (4-1)/(0-1) which equals 3/-1. 3/-1 is your slope. So, your answer in slop intercept form is: y= 3/-1x+4. You could also try y= -3x+4 if that makes you more comfortable.

I know this is long this is my first time doing this lol.

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Simplify the leading term as

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Quotient rule:

$\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'=\dfrac{(1+\sin(2x))(\cos(2x))'-\cos(2x)(1+\sin(2x))'}{(1+\sin(2x))^2}$

Chain rule:

$(\cos(2x))'=-\sin(2x)(2x)'=-2\sin(2x)$

$(1+\sin(2x))'=\cos(2x)(2x)'=2\cos(2x)$

Put everything together and simplify:

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$=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2}{(1+\sin(2x))^2}$

$=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac{\sin(2x)+1}{(1+\sin(2x))^2}$

$=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac1{1+\sin(2x)}$

$=-\dfrac1{\sqrt{\cos(2x)}}\dfrac1{\sqrt{1+\sin(2x)}}$

$=\boxed{-\dfrac1{\sqrt{\cos(2x)(1+\sin(2x))}}}$

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