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Monica [59]
3 years ago
8

A company has two packaging machines in a unit, each with a different daily capacity. The capacity of machine 1 is defined by th

e function f(m) = (m + 4)2 + 100, and the capacity of machine 2 is defined by the function g(m) = (m + 12)2 − 50, where m is the number of minutes the packaging machine operates. Create the function C(m) that represents the combined capacity of the two machines.
Mathematics
1 answer:
ollegr [7]3 years ago
6 0

Total capacity = sum of the individual production capacities.

Here,

Total capacity = sum of f(m) = (m + 4)^2 + 100  and g(m) = (m + 12)^2 − 50.

Then f(m) + g(m) =  (m + 4)^2 + 100 + (m + 12)^2 − 50.

We must expand the binomial squares in order to combine like terms:

 m^2 + 8 m + 16 + 100

+m^2 + 24m + 144 - 50

---------------------------------

Then f(m) + g(m) = 2m^2 + 32m + 160 + 50  

      f(m) + g(m)   = 2m^2 + 32m + 210, where m is the number of

                                minutes during which the two machines operate.

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Masja [62]

Answer:

One number is 17 and the other is 85.

Step-by-step explanation:

Let the numbers be m and n.  Let m = (1/5)n.  Note that m+n = 18.

Substituting (1/5)n for m in the 2nd equation, we get (1/5)n + (5/5)n = 18.  Multiplying all three terms by 5 to remove fractions, we get n + 5 = 90, so that n = 85.  m is (1/5) of 85, or m = 17.

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2 years ago
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Artist 52 [7]

Answer:

Step-by-step explanation:

Distance traveled in 1\frac{4}{5} hours = \frac{3}{10}

Distance traveled in 1 hour = \frac{3}{10} ÷ 1\frac{4}{5}

          =\frac{3}{10} ÷ \frac{9}{5}

          = \frac{3}{10}*\frac{5}{9}\\\\=\frac{1}{2}*\frac{1}{3}\\\\=\frac{1}{6}

distance traveled in 3 1/5 hours = \frac{1}{6}*3\frac{1}{5}

 ==\frac{1}{6}*\frac{8}{5}\\\\=\frac{1}{3}*\frac{4}{5}\\\\=\frac{4}{15}

5 0
2 years ago
What is the point of intersection when the system of equations below is graphed on the coordinate plane?
lora16 [44]

Answer:

<h2>not exist</h2>

Step-by-step explanation:

The coordinates of the intersection of the line are the solution of the system of equations.

\underline{+\left\{\begin{array}{ccc}x-y=1\\y-x=1\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad\qquad0=2\qquad\bold{FALSE}

The system of equations has no solution. Therefore, the lines are parallel (the intersection does not exist).

5 0
3 years ago
Help needed thanks it greatly appretiated
Ivan

c is the correct answer of this question

5 0
2 years ago
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Check the picture below

it can't be -3... since is width unit, so it has to be the other

and recall that length = 2w - 5

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3 years ago
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