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Keith_Richards [23]
3 years ago
9

The average density of Styrofoam is 1.00 kg/m^3. If a Styrofoam cooler is made with outside dimensions of 50.0 x 35.0 x 30.0 cm

and the uniform thickness of the Styrofoam is 3.00 cm (including the lid), what is the volume of the Styrofoam used in cubic inches? The answer should be 1100 cubic inches, but I do not know how to get there

Mathematics
1 answer:
Nadusha1986 [10]3 years ago
5 0

Answer:

View graph

Step-by-step explanation:

The closest by default to 1100 cubic inches is taking two side and two front covers and that would give you 1080.12 cubic inches since two of the covers you must subtract 3 cm per side and side which would add 6 cm in total so outside would be 35 cm and indoor 29 cm . According the graph

29*50*3= 4350 * 2 lid = 8700 cm3

30*50*3= 4500*2 lid = 9000 cm3

So 8700+9000 = 17700 cm3 to inches 3 = 1080.12 cm3

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Verify that the points are the vertices of a parallelogram and find its area. (2,-1,1), (5, 1,4), (0,1,1), (3,3,4)
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Answer:

Area = 13.15 square units

Step-by-step explanation:

Let the given vertices be represented as follows:

A(2, -1, 1) = 2i - j + k

B(5, 1, 4) = 5i + j + 4k

C(0, 1, 1) = 0i + j + k

D(3, 3, 4) = 3i + 3j + 4k

(i) Let's calculate the vectors of all the sides:

\\AB = B - A =  (5i + j + 4k) - (2i - j + k)

AB = 5i + j + 4k - 2i + j - k                 [Collect like terms]

AB = 3i + 2j + 3k

BC = C - B =  (0i + j + k) - (5i + j + 4k)

BC = 0i + j + k - 5i - j - 4k                 [Collect like terms]

BC = -5i + 0j - 3k

CD = D - C =  (3i + 3j + 4k) - (0i + j + k)

CD = 3i + 3j + 4k - 0i - j - k                [Collect like terms]

CD = 3i + 2j + 3k

DA = A - D =  (2i - j + k) - (3i + 3j + 4k)

DA = 2i - j + k - 3i - 3j - 4k                [Collect like terms]

DA = -i - 4j - 3k

AC = C - A =  (0i + j + k) - (2i - j + k)

AC = 0i + j + k - 2i + j - k                [Collect like terms]

AC = -2i + 2j

BD = D - B = (3i + 3j + 4k) - (5i + j + 4k)

BD = 3i + 3j + 4k - 5i - j - 4k                [Collect like terms]

BD = -2i + 2j

(ii) From the results in (i) above, we can deduce that;

AB = CD This implies that AB || CD  [AB is parallel to CD]

AC = BD This implies that AC || BD  [AC is parallel to BD]

(iii) Therefore, ABDC is a parallelogram since opposite sides (AB and CD) are parallel. Hence, the points are vertices of a parallelogram

<u>Now let's calculate the area</u>

To find the area of the parallelogram, we find the magnitude of the cross product of any two adjacent sides.

In this case, we'll choose AB and AC

Area = |AB X AC|

Where;

AB X AC = \left[\begin{array}{ccc}i&j&k\\3&2&3\\-2&2&0\end{array}\right]

<u></u>

AB X AC = i(0 - 6) - j(0 + 6) + k(6 + 4)

AB X AC = - 6i - 6j + 10k

|AB X AC| = \sqrt{(-6)^2 + (-6)^2 + (10)^2}

|AB X AC| = \sqrt{172}

|AB X AC| = 13.15

Area = 13.15 square units.

<u></u>

<u></u>

<u>PS: </u> ACBD is also a parallelogram. The diagram has also been attached to this response.

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