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strojnjashka [21]
3 years ago
6

Hideki is draining a 45-gallon aquarium at a rate of 15 gallons per minute. What is the initial value of the linear function tha

t represents this situation?
Mathematics
2 answers:
kkurt [141]3 years ago
8 0
3 min sence 15 will enter 45 3 times
Alexxx [7]3 years ago
3 0

Answer:

Hideki is draining a 45-gallon aquarium at a rate of 15 gallons per minute. Let's solve for the initial value of the linear function that represents thissituation.

=> 45 gallon aquarium at rate of 15 gallons per minute.

=> 45 gallon / 15 gallon = 3 minutes.

Therefore with the rate of 15 gallons per minutes, Hideki will be able to finish the 45 gallon aquarium in 3 minutes.

Step-by-step explanation:

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gladu [14]

Answer: x = 11/20 = 0.550

Step-by-step explanation:

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                    x-2/5-(3/20)=0  

Step by step solution :

Step  1  :

            3

Simplify   ——

           20

Equation at the end of step  1  :

       2      3

 (x -  —) -  ——  = 0  

       5     20

Step  2  :

           2

Simplify   —

           5

Equation at the end of step  2  :

       2      3

 (x -  —) -  ——  = 0  

       5     20

Step  3  :

Rewriting the whole as an Equivalent Fraction :

3.1   Subtracting a fraction from a whole

Rewrite the whole as a fraction using  5  as the denominator :

         x     x • 5

    x =  —  =  —————

         1       5  

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

3.2       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:

x • 5 - (2)     5x - 2

———————————  =  ——————

     5            5    

Equation at the end of step  3  :

 (5x - 2)     3

 ———————— -  ——  = 0  

    5        20

Step  4  :

Calculating the Least Common Multiple :

4.1    Find the Least Common Multiple

     The left denominator is :       5  

     The right denominator is :       20  

       Number of times each prime factor

       appears in the factorization of:

Prime  

Factor   Left  

Denominator   Right  

Denominator   L.C.M = Max  

{Left,Right}  

5 1 1 1

2 0 2 2

Product of all  

Prime Factors  5 20 20

     Least Common Multiple:

     20  

Calculating Multipliers :

4.2    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 = 1

Making Equivalent Fractions :

4.3      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.      (5x-2) • 4

  ——————————————————  =   ——————————

        L.C.M                 20    

  R. Mult. • R. Num.       3

  ——————————————————  =   ——

        L.C.M             20

Adding fractions that have a common denominator :

4.4       Adding up the two equivalent fractions

(5x-2) • 4 - (3)     20x - 11

————————————————  =  ————————

       20               20    

Equation at the end of step  4  :

 20x - 11

 ————————  = 0  

    20    

Step  5  :

When a fraction equals zero :

5.1    When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

 20x-11

 —————— • 20 = 0 • 20

   20  

Now, on the left hand side, the  20  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :

  20x-11  = 0

Solving a Single Variable Equation :

5.2      Solve  :    20x-11 = 0  

Add  11  to both sides of the equation :  

                     20x = 11

Divide both sides of the equation by 20:

                    x = 11/20 = 0.550

One solution was found :

                  x = 11/20 = 0.550

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