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MissTica
3 years ago
12

A toy company intends to ship out its entire stock of 10,254 toy trucks, and three times as many toy cars. If each box holds 35

toys of either type, how many full boxes can be packed, and how many toys are left over for the last box?
Mathematics
1 answer:
svetoff [14.1K]3 years ago
6 0
There are 10,254 toy trucks

and 10,254 * 3 = 30,762 toy cars.

In total, there are 10,254+30,762= 41,016 toys 

when we divide 41,016 by 35, the quotient is the number of the full boxes, and the remainder is the number of toys left.

41,016=35*1171+31


Answer: 1171 full boxes, 31 toys left for the last box
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Answer:

The bounded area is 5 + 5/6 square units. (or 35/6 square units)

Step-by-step explanation:

Suppose we want to find the area bounded by two functions f(x) and g(x) in a given interval (x1, x2)

Such that f(x) > g(x) in the given interval.

This area then can be calculated as the integral between x1 and x2 for f(x) - g(x).

We want to find the area bounded by:

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To find this area, we need to f(x) - g(x) between x = -1 and x = 2

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We know that:

\int\limits^{}_{} {x} \, dx = \frac{x^2}{2}

\int\limits^{}_{} {1} \, dx = x

\int\limits^{}_{} {x^2} \, dx = \frac{x^3}{3}

Then our integral is:

\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx = (\frac{2^3}{2}  + 2 - \frac{2^2}{2}) - (\frac{(-1)^3}{3}  + (-1) - \frac{(-1)^2}{2}  )

The right side is equal to:

(4 + 2 - 2) - ( -1/3 - 1 - 1/2) = 4 + 1/3 + 1 + 1/2 = 5 + 2/6 + 3/6 = 5 + 5/6

The bounded area is 5 + 5/6 square units.

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