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Genrish500 [490]
3 years ago
8

Awnsers pleasesolve it

Mathematics
1 answer:
sukhopar [10]3 years ago
5 0

1)30

2)700

3)7

4)9 000

5)30 000

6)500

7)800 000

8)6 000

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4523.89m

So for a cylinder it is radius times two. Then multiply your answer by pie 3.14 then times high.

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Decrease £16855.21 by 13.5% <br> Give your answer rounded to 2DP
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Jose has 4 friends and 3 apples how much would each friend get if he shared
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\bf \cfrac{\stackrel{\textit{total amount of apples}}{3}}{\stackrel{\textit{total amount of people}}{4}}\qquad \qquad \quad\stackrel{Jose}{ \cfrac{3}{4}}+\stackrel{friend}{\cfrac{3}{4}}+\stackrel{friend}{\cfrac{3}{4}}+\stackrel{friend}{\cfrac{3}{4}}=\cfrac{12}{4}=3

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Solve -5x2 - 20x + 35 = 30 by completing the square.
Semenov [28]

Answer:

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Step-by-step explanation:

-5x2 - 20x + 35 = 30

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5 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

4 0
3 years ago
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