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wariber [46]
2 years ago
8

What is the difference of -9 + 2i and 4 - 6i?

Mathematics
1 answer:
MAVERICK [17]2 years ago
8 0

Answer:

-13-4i

Step-by-step explanation:

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The Basketball Boosters Club plans a pancake supper to raise $1,200. They can serve 240 people. They
wolverine [178]

\huge{ \mathfrak{  \underline{ Answer} \:  \:  ✓ }}

let the number of

  • adults be x

  • children be y

Total number of people : adults + children

  • x + y = 240

  • x = 240 - y - [1]

Total money earned from adults :

  • 6 × x = 6x

Total money earned from children :

  • 2 × y = 2y

Therefore, total money = money earned from each Adult and children.

  • 6x + 2y = 1200 -[2]

Now, let's plug the value of x from equation [1] into equation [2]

  • 6 (240 - y) + 2y = 1200

  • 1440 - 6y + 2y = 1200

  • -4y = 1200 - 1440

  • -4y = - 240

  • y = 60

let's plug the value of y in equation [1]

  • x = 240 - y

  • x = 240 - 60

  • x = 180

Hence, we get :

  • total adults = x = 180

  • total children = y = 60

Correct option is C

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\mathrm{ ☠ \: TeeNForeveR \: ☠}

7 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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