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Dovator [93]
3 years ago
14

Please help with this

Mathematics
1 answer:
olya-2409 [2.1K]3 years ago
8 0
Theyre similar so answer is B
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What is 12 times 164
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Answer:

1968

Step-by-step explanation:

give me brainliest plsssss tysm

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There are 8 watermelons and 6 apples in the shopping cart. What is the ratio of
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<h2><u>RATIO</u></h2>

<h3><u>\underline \mathcal{PROBLEM:}</u></h3>

» There are 8 watermelons and 6 apples in the shopping cart. What is the ratio of apples to total fruits? Be sure to write your answer in lowest terms.

<h3>\underline{ \mathcal{ANSWER:}}</h3>
  • \blue{ \bold{4:7}}

— — — — — — — — — —

<h3>\underline{ \mathcal{EXPLANATION:}}</h3>

Based on the problem, there are 8 watermelons and 6 apples in the shopping cart. So,

  • \sf \: Apple =  \green6

  • \sf \: Watermelon = \green 8

  • \sf \: Total \:  Fruits= \green {14 }

Therefore, the ratio of apples to total fruits is 8:14 and its lowest term is 4:7.

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3 0
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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