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kykrilka [37]
3 years ago
15

A video game that usually costs $30.65 is marked down 60%. Kelvin determined that the new price of the game would be $18.39. Loo

k at Kelvin’s work and find his error.
($30.65)(0.60) = $18.39
Mathematics
3 answers:
nata0808 [166]3 years ago
8 0
60% of 30.65 is $18.39
Kelvin thought that was the final number but no so you now must subtract $30.65 from $18.39 which is then your final answer $12.39
otez555 [7]3 years ago
6 0

Answer:

Kelvin wrote 60 the wrong way.

He has to write $30.65 - 60%, not 0.60.

Here is the final correct equation:

$30.65 - 60% = 12.26

denil2 years ago
1 0

Sample Response: Kelvin should have subtracted $18.39 from $30.65 to find the marked-down price. He could also have multiplied the original cost by 40% to find the sale price of $12.26.

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Make this value negative. The reason why is due to the fact that we have a sort of reflection going on as we're scaling down the figure. 

So the final answer is  -0.2

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The shadow of a vertical tower is 72.0-ft long when the angle of elevation of the
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38.15 ft is the height of the tower using tangent
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3 years ago
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

 =>       d\theta_1  = 0.25 p

Converting to degree

           d\theta_1  = (0.25* 57.29) p

            d\theta_1  =(14.36p)^o

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morpeh [17]

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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