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drek231 [11]
3 years ago
12

At a speed of 45 yards per minute, a 120 pound swimmer burns 420 calories per hour and a 172 pound swimmer burns 600 calories pe

r hour. write an equation that models the number of calories burned per hour as a function of the swimmer's weight.
Mathematics
1 answer:
Murrr4er [49]3 years ago
3 0
<span>c = w*45/13 + 60/13 I will be assuming that the equation will be linear since there are only 2 data points available. The form of the equation will be c = aw + b where c = calories burned a = some constant w = weight b = some other constant. If you look at the equation, you'll notice that it looks like the equation for a line i slope intercept form. So first, let's calculate the slope of the desired equation. a = (600-420)/(172-120) = 180/52 = 45/13 = 3.461538462 Now we have the equation c = w*45/13 + b Substitute the values for a known point, giving 420 = 120*45/13 + b Now solve for b 420 = 120*45/13 + b 420 = 5400/13 + b 420 - 5400/13 = b 5460/13 - 5400/13 = b 60/13 = b So now the equation is c = w*45/13 + 60/13</span>
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*ੈ✩‧₊˚<br> raise the fraction to higher terms as indicated <br><br> 2/3 to twenty-firsts
DIA [1.3K]

The appropriate fraction when the number is raised to higher terms will be 14/21.

<h3>How to compute the fraction?</h3>

It should be noted that when we sat raising that fraction to higher terms, it implies that changing the numerator and the denominator to higher numbers to reflect the fraction.

Here, the given number is 2/3. When it's raises higher, the appropriate fraction will be:

2/3 = x/21

x = (2 × 21)/3

x = 42/3

x = 14

Therefore, the fraction will be 14/21.

Learn more about fractions on:

brainly.com/question/17220365

#SPJ1

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Elaine Mellon earns $17.80 per hour as an electric billing specialist for a legal clinic. How many hours per week must she work
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Step-by-step explanation:

8 0
3 years ago
Better Products, Inc., manufactures three products on two machines. In a typical week, 40 hours are available on each machine. T
Kaylis [27]

Answer:

z (max)  =  1250 $

x₁  = 25    x₂  =  0   x₃  =  25

Step-by-step explanation:

                                Profit $    mach. 1      mach. 2

Product 1     ( x₁ )       30             0.5              1

Product 2    ( x₂ )       50             2                  1

Product 3    ( x₃ )       20             0.75             0.5

Machinne 1 require  2 operators

Machine   2 require  1  operator

Amaximum of  100 hours of labor available

Then Objective Function:

z  =  30*x₁  +  50*x₂  +  20*x₃      to maximize

Constraints:

1.-Machine 1 hours available  40

In machine 1    L-H  we will need

0.5*x₁  +  2*x₂  + 0.75*x₃  ≤  40

2.-Machine 2   hours available  40

1*x₁  +  1*x₂   + 0.5*x₃   ≤  40

3.-Labor-hours available   100

Machine 1     2*( 0.5*x₁ +  2*x₂  +  0.75*x₃ )

Machine  2       x₁   +   x₂   +  0.5*x₃  

Total labor-hours   :  

2*x₁  +  5*x₂  +  2*x₃  ≤  100

4.- Production requirement:

x₁  ≤  0.5 *( x₁ +  x₂  +  x₃ )     or   0.5*x₁  -  0.5*x₂  -  0.5*x₃  ≤ 0

5.-Production requirement:

x₃  ≥  0,2 * ( x₁  +  x₂   +  x₃ )  or    -0.2*x₁  - 0.2*x₂ + 0.8*x₃   ≥  0

General constraints:

x₁  ≥   0       x₂    ≥   0       x₃     ≥   0           all integers

The model is:

z  =  30*x₁  +  50*x₂  +  20*x₃      to maximize

Subject to:

0.5*x₁  +  2*x₂  + 0.75*x₃  ≤  40

1*x₁  +  1*x₂   + 0.5*x₃       ≤  40

2*x₁  +  5*x₂  +  2*x₃        ≤  100

0.5*x₁  -  0.5*x₂  -  0.5*x₃  ≤ 0

-0.2*x₁  - 0.2*x₂ + 0.8*x₃   ≥  0

x₁  ≥   0       x₂    ≥   0       x₃     ≥   0           all integers

After 6 iterations with the help of the on-line solver AtomZmaths we find

z (max)  =  1250 $

x₁  = 25    x₂  =  0   x₃  =  25

6 0
3 years ago
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