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ale4655 [162]
3 years ago
9

Someone please help me!

Mathematics
1 answer:
balandron [24]3 years ago
3 0

Answer:

The answer to your question is below

Step-by-step explanation:

43.-

x = \sqrt{27^{2} + 22^{2}- 2(27)(22)cos 73}

x = \sqrt{865.66}

x = 29.42

44.-

x = \sqrt{10^{2} + 14^{2} -2(10)(14)cos 66}

x = \sqrt{182.11}

x = 13.49

45.-

cos x = \frac{11^{2} - 17^{2} - 10^{2}}{-2(11)(17)}

cos x = 0.7166

     x = 44.22°

46.-

x = \sqrt{16^{2} + 12^{2} -2(16)(12)cos75}

x = \sqrt{300.61}

x = 17.34

47.-

cos P = \frac{6^{2} - 13^{2} -11^{2}}{-2(13)(11)}

cos P = 0.888

     P = 27.36°

sinR/13 = sinP/6

sin R = 13sin27.36/6

sinR = 0.996

   R = 84.71°

Q = 180 - 84.71 - 27.36

Q = 67.93°

48.-

D = 180 - 25 - 113

D = 42°

CD = 9Sin113/sin42

CD = 12.38

ED = 9sin25/sin42

ED = 5.68

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Given parameters;

  Let us solve this problem step by step;

Let us represent Simon's money by S

Kande's money by K

  • Simon has more money than Kande

               S > K

  •  if Simon gave Kande K20, they would have the same amount;

 if Simon gives $20, his money will be  S - 20 lesser;

      When Kande receives $20, his money will increase to K + 20

                     S - 20  = K + 20   ------ (i)

  • While if Kande gave Simon $22, Simon would then have twice as much as Kande;

          if Kande gave Simon $22, his money will be K - 22

    Simon's money, S + 22;

                  S + 22  = 2(K - 22)    ------ (ii)

Now we have set up two equations, let us solve;

         S - 20  = K + 20  ---- i

         S + 22  = 2(K - 22)  ;       S + 22  = 2K - 44  ---- ii

So,      S - 20  = K + 20

          S + 22  = 2K - 44

subtract both equations;

               -20 - 22  = (k -2k)  + 64

                   -42  = -k + 64

                       k  = 106

Using equation i, let us find S;

            S - 20 = K + 20

             S - 20  = 106 + 20

              S = 106 + 20 + 20  = 146

Therefore, Kande has $106 and Simon has $146

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