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Vera_Pavlovna
9 days ago
6

A contractor can spend at most $350 a day on operating costs and payroll. It costs $75 each day to operate the forklift and $55

a day for each crew member. Write an inequality to represent the contractor's budget for the day.
75 + 55x ≥ 350


75 + 55x ≤ 350


75x + 55 ≥ 350


75x + 55 ≤ 350

I think it's the second one, but I'm not sure!
Mathematics
1 answer:
tester [12.3K]9 days ago
8 0

75 + 55x ≤ 350

Step-by-step explanation:

  • Dado que no conocemos el número de trabajadores que se les paga $55 por día, asignamos el valor desconocido como x. Entonces, el total que se está pagando a los trabajadores es;

55x

  • El total gastado, en $, por día (incluyendo el costo de operación) será;

75 + 55x

  • Según la ecuación, el máximo que el contratista puede gastar en un día en costos operativos y nómina es de $350. Por lo tanto, 75 + 55x no puede exceder 350 pero puede ser igual o menor que;

75 + 55x ≤ 350

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10 times as many as blank hundreds or 60 hundreds is blank thousands
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The result of multiplying 10 by 6 hundreds equals 60 hundreds, which can also be expressed as 6 thousands.

Explanation

This can be simplified as follows:

  • 10 times a given quantity of _____ hundreds results in 60 hundreds
  • 60 hundreds convert to _____ thousands

We will divide this into two segments:

\boxed{ \ 10 \times (N \times 100) = 60 \times 100 \ }... Equation-1

\boxed{ \ 60 \times 100) = M \times 1,000 \ }... Equation-2

This straightforward multiplication relates to the place value system. We need to perform calculations to find the values of N and M.

Initially, let's focus on Equation-1.

We will move the variable M to one side of the equation in order to isolate it and solve for its value.

\boxed{ \ 10 \times (N \times 100) = 60 \times 100 \ }

\boxed{ \ 1,000 \times N = 6,000 \ }

Both parts will be divided by 1,000, essentially being multiplied by \frac{1}{1,000}.

\boxed{ \ \frac{1}{1,000} \times 1,000 \times N = \frac{1}{1,000} \times 6,000 \ }

Thus, we conclude with \boxed{\boxed{ \ N = 6 \ }}

Next, we process Equation-2 to derive M's value.

\boxed{ \ 60 \times 100) = M \times 1,000 \ }

\boxed{ \ 6,000 = M \times 1,000 \ }

Rearranging this equation to place M on the left side appears as follows:

\boxed{ \ M \times 1,000 = 6,000 \ }

Again, both sides undergo division by 1,000, which translates to multiplication by \frac{1}{1,000}..

\boxed{ \ \frac{1}{1,000} \times M \times 1,000 = \frac{1}{1,000} \times 6,000 \ }

This results in \boxed{\boxed{ \ M = 6 \ }}

- - - - - - -

Alternative approach for the second step:

60 hundreds equals to __ thousands

\boxed{ \ 6 \times 10 \ hundreds = \ M \ thousands} \ }

\boxed{ \ 6 \times thousands = \ M \ thousands} \ }

\boxed{\boxed{ \ M = 6 \ }}

Additional resources

  1. A more detailed version of this topic
  2. 2 thousands 7 tens divided by 10 equals what?
  3. 100 is equivalent to 1/10 of which number?

Keywords: 10 times as many as, blank, hundreds, 60 hundreds, or, thousands, isolate, operations, multiply, divide, fraction, both sides, equal, the opposite, both sides are multiplied by, divided by, rearrange

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