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monitta
3 years ago
6

The equation y minus two-thirds = negative 2 (x minus one-fourth) is written in point-slope form. What is the equation in slope-

intercept form? Y = negative 2 x minus StartFraction 1 Over 6 EndFraction y = negative 2 x + StartFraction 7 Over 6 EndFraction y = negative 2 x + StartFraction 5 Over 12 EndFraction y = negative 2 x minus StartFraction 11 Over 12 EndFraction
Mathematics
2 answers:
victus00 [196]3 years ago
7 0

Answer:

b

Step-by-step explanation:

Lena [83]3 years ago
5 0

Answer:

(B)  y=-2x+ \dfrac{7}{6}

Step-by-step explanation:

Given the point-slope form of the equation of a line below:

y-\dfrac{2}{3}=-2\left(x- \dfrac{1}{4}\right)

We are required to write it in the slope-intercept form, y=mx+c.

y-\dfrac{2}{3}=-2\left(x- \dfrac{1}{4}\right)\\\\y-\dfrac{2}{3}=-2x+ \dfrac{2}{4}\\\\y=-2x+ \dfrac{2}{4}+\dfrac{2}{3}\\\\y=-2x+ \dfrac{7}{6}

The slope intercept form is:

y=-2x+ \dfrac{7}{6}

<u>The correct option is B.</u>

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You stand a known distance from the base of the tree, measure the angle of elevation the top of the tree to be 15â—¦ , and then
gogolik [260]

Answer:

The maximum possible error of in measurement of the angle is  d\theta_1  =(14.36p)^o

Step-by-step explanation:

From the question we are told that

    The angle of elevation  is  \theta_1  =  15 ^o =  \frac{\pi}{12}

     The height of the tree is  h

      The distance from the base is  D

h is mathematically represented as

            h  = D tan \theta       Note : this evaluated using SOHCAHTOA i,e

                                               tan\theta  =  \frac{h}{D}

Generally for small angles the series approximation of  tan \theta \  is

          tan \theta  =  \theta  + \frac{\theta ^3 }{3}

So given that \theta =  15 \ which \ is \ small

       h = D (\theta + \frac{\theta^3}{3} )

       dh = D (1 + \theta^2) d\theta

=>        \frac{dh}{h} =  \frac{1 + \theta ^2}{\theta + \frac{\theta^3}{3} } d \theta

Now from the question the relative error of height should be at  most

        \pm  p%

=>    \frac{dh}{h} =   \pm p

=>    \frac{1 + \theta ^2}{\theta + \frac{\theta^3}{3} } d \theta  = \pm p

=>      d\theta  =  \pm  \frac{\theta +  \frac{\theta^3}{3} }{1+ \theta ^2} *    \ p

 So  for   \theta_1

            d\theta_1  =  \pm  \frac{\theta_1 +  \frac{\theta^3_1 }{3} }{1+ \theta_1 ^2} *    \ p

substituting values  

          d [\frac{\pi}{12} ]  =  \pm  \frac{[\frac{\pi}{12} ] +  \frac{[\frac{\pi}{12} ]^3 }{3} }{1+ [\frac{\pi}{12} ] ^2} *    \ p

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Converting to degree

           d\theta_1  = (0.25* 57.29) p

            d\theta_1  =(14.36p)^o

4 0
3 years ago
I need help on this one please
zepelin [54]

Answer:

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3 years ago
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Step-by-step explanation:

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7 0
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