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raketka [301]
3 years ago
8

A motorcyclist leaves City A and rides at a constant speed towards City B, which is 225 km away. After 1.5 hours, the motorcycli

st stops a half an hour lunch. To reach City B on time, the motorcyclist increases his speed after lunch by 10km/hour. Find the motorcyclists original speed
Mathematics
2 answers:
rewona [7]3 years ago
5 0

Answer:

50 km/hr

Step-by-step explanation:

OK, this is how I solved this problem.

Let r = original rate of speed

     t = time after the stop to finish the trip

Time for trip without stopping is 225/r

(1)     1.5 + .5 + t = 225/r      This is a time equation

(2)    1.5r + t(r + 10) = 225     This is a distance equation

(1)       2 + t = 225/r            (2)      1.5r + tr + 10t = 225

         2r + tr = 225

                tr = 225 - 2r                 1.5r + 225 - 2r + 10(225/r - 2) = 225

                t = 225/r - 2

                                                   -.5r + 225 + 2250/r - 20 = 225

                                                     -5r + 2250 + 22500/r - 200 = 2250

                                                     -5r^2 + 2250r + 22500 - 200r = 2250r

                                                       5r^2 + 200r - 22500 = 0

                                                       5(r^ + 40r - 4500) = 0

                                                           5(r - 50)(r + 90) = 0

                                                                 r = 50   or    r = -90

The rate cannot be negative, so the original speed 50 km/hr

This not an easy problem and I hope you were able to follow my work

Rudiy273 years ago
4 0

Answer:50km hope that helped pls mark me brainliest :))

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djverab [1.8K]

Answer:

  -25            

 ——— = -2.08333

    12  

Step-by-step explanation:

Step  1  :

           7

Simplify   —

           4

Equation at the end of step  1  :

       1           7

 (0 -  —) +  (0 -  —)

       3           4

Step  2  :

           1

Simplify   —

           3

Equation at the end of step  2  :

       1     -7

 (0 -  —) +  ——

       3     4

Step  3  :

Calculating the Least Common Multiple :

3.1    Find the Least Common Multiple

     The left denominator is :       3

     The right denominator is :       4

       Number of times each prime factor

       appears in the factorization of:

Prime

Factor   Left

Denominator   Right

Denominator   L.C.M = Max

{Left,Right}

3 1 0 1

2 0 2 2

Product of all

Prime Factors  3 4 12

     Least Common Multiple:

     12

Calculating Multipliers :

3.2    Calculate multipliers for the two fractions

   Denote the Least Common Multiple by  L.C.M

   Denote the Left Multiplier by  Left_M

   Denote the Right Multiplier by  Right_M

   Denote the Left Deniminator by  L_Deno

   Denote the Right Multiplier by  R_Deno

  Left_M = L.C.M / L_Deno = 4

  Right_M = L.C.M / R_Deno = 3

Making Equivalent Fractions :

3.3      Rewrite the two fractions into equivalent fractions

Two fractions are called equivalent if they have the same numeric value.

For example :  1/2   and  2/4  are equivalent,  y/(y+1)2   and  (y2+y)/(y+1)3  are equivalent as well.

To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier.

  L. Mult. • L. Num.      -1 • 4

  ——————————————————  =   ——————

        L.C.M               12  

  R. Mult. • R. Num.      -7 • 3

  ——————————————————  =   ——————

        L.C.M               12  

Adding fractions that have a common denominator :

3.4       Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

-1 • 4 + -7 • 3     -25

———————————————  =  ———

      12            12

Final result :

 -25            

 ——— = -2.08333

 12            

Processing ends successfully

plz mark me as brainliest :)

7 0
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Step-by-step explanation:

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