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3 years ago

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One warehouse had 185 tons of coal, another one had 237 tons. The first warehouse delivered 15 tons of coal to its clients daily

, the second one delivered 18 tons of coal to its clients daily. In how many days will the second warehouse have 1.5 times more coal than the first one?

1 answer:

lianna [129]3 years ago

7 0

Answer:

In 9 days will the second warehouse have 1.5 times more coal than the first one

Step-by-step explanation:

The coal left in warehouses after x days can be found using the equation:

first warehouse: 185 - 15x

second warehouse= 237-18x

Let the second warehouse have 1.5 times more coal than the first one after n days then

1.5 ×(185 - 15n)=237-18n which gives:
277,5-22,5n=237-18n and
40,5=4,5n
9=n

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Step-by-step explanation:

The Perimeter of the full shape is the sum of the lengths of the edges of the parts. For convenience in referencing them, we'll call the large curve " $curve_{big}$ " and the three smaller curves " $curve_1$ " " $curve_2$ " " $curve_3$ " in order from left to right.

Thus, the Perimeter of the full shape can be written as an equation:

$P_{overall} = Length(curve_{big})+Length(curve_1)+Length(curve_2)+Length(curve_3)$ Since all of those edge lengths are curves, and the question states that all of the curves are made from parts of circles, then we need to know how to find the length of the edge of a circle.

<u>Parts of a circle</u>

Since values in the diagram are diameters, use the formula for the Perimeter of a circle $P=\pi d$ (where d is the diameter).

Let's call the diameters of each of our curves " $d_{big}$ " " $d_1$ " " $d_2$ " " $d_3$ ", with the subscripts denoting which curve we're referring to.

Note that for each curve, the curve only represents half of a circle. So, to find the length of each curve, we'll need half of the full perimeter of each circle.

So for instance: $Length(curve_{big})=\frac{1}{2} \pi d_{big}$

Substituting back into the main equation above:

$P_{overall} = Length(curve_{big})+Length(curve_1)+Length(curve_2)+Length(curve_3)$ $P_{overall}=\frac{1}{2} \pi d_{big} + \frac{1}{2} \pi d_{1} + \frac{1}{2} \pi d_{2} + \frac{1}{2} \pi d_{3}$

Note that all terms have common factors of "one-half" and "pi" in them. These can be factored out:

$P_{overall}=\frac{1}{2} \pi (d_{big} + d_{1} + d_{2} +d_{3})$

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Substituting and simplifying (in terms of pi):

$P_{overall}=\frac{1}{2} \pi ( (12cm) + (4cm) + (4cm) + (4cm) )\\P_{overall}=\frac{1}{2} \pi ( 24cm)\\P_{overall}=12 \pi cm$

<u>Additional Understanding</u>

Interesting for this problem, since the diameters of the 3 small curves formed the diameter of the large curve $d_{1} + d_{2} + d_{3} =d_{big}$ , one could make a different substitution into one of our formulas above:

$P_{overall}=\frac{1}{2} \pi (d_{big} + d_{1} + d_{2} +d_{3})$

$P_{overall}=\frac{1}{2} \pi (d_{big} + (d_{big}))$

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