In 1999, Marks & Spencer, a British department store, created the biggest sandwich ever made. The tuna and cucumber sandwich was in the form of a triangular prism. Suppose each slice of bread was 8 inches thick. Find each measure to the nearest tenth. a. The surface area in square feet of the sandwich when filled b. The volume of filling in cubic feet to the nearest tenth

Answers

Answer 1

Answer:

A) 19.2 ft^2

B) 28.9 ft^3

Step-by-step explanation:

Thickness = 8 inches

Assuming the triangular prism is made up of an equilateral triangle hence all sides are equal

lets assume the side = 10 inches  

lets assume the thickness as the height of the prism = 8 inches

Assume length of the rectangle(l) = 5 feet

A) The surface area of the sandwich triangular prism

total surface area = bh + 2ls + lb

                             = 10*8 + 2(5*10) + ( 5 * 10 )

                             = 80 + 100 + 50 = 230 inches = 19.2 ft^2

B ) The volume of filing in cubic feet to the nearest tenth

V = [tex]\frac{1}{4} h \sqrt{-a^4 +2(ab)^2+2(ac)^2-b^4+2(bc)^2-c^4}[/tex]  ------- 1

where : a = b = c = 10 inches

h = 8 inches

input the given values into the equation 1 above

V= 346.41 inches ^3  =  28.9 ft^3

 


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[tex]

\underline{\bf{Given\::}}

Given:

\underline{\bf{To\:find\::}}

Tofind:

\underline{\bf{Explanation\::}}

Explanation:

\boxed{\bf{\frac{1}{f} =\frac{1}{v} -\frac{1}{u} }}}}

\begin{gathered}\longrightarrow\sf{\dfrac{1}{-10} =\dfrac{1}{v} -\dfrac{1}{-30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{1}{-10} +\dfrac{1}{30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{-3+1}{30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\cancel{\dfrac{-2}{30} }}\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{1}{-15} }\\\\\\\longrightarrow\sf{v=-15\:cm}\end{gathered}

−10

1

=

v

1

−30

1

v

1

=

−10

1

+

30

1

v

1

=

30

−3+1

v

1

=

30

−2

v

1

=

−15

1

⟶v=−15cm

\boxed{\bf{M \:A \:G \:N\: I \:F \:I \:C\: A\: T \:I \:O\: N :}}

MAGNIFICATION:

\begin{gathered}\mapsto\sf{m=\dfrac{Height\:of\:image\:(I)}{Height\:of\:object\:(O)} =\dfrac{Distance\:of\:image}{Distance\:of\:object} =\dfrac{v}{u} }\\\\\\\mapsto\sf{m=\cancel{\dfrac{-30}{-15}} }\\\\\\\mapsto\bf{m=2\:cm}\end{gathered}

↦m=

Heightofobject(O)

Heightofimage(I)

=

Distanceofobject

Distanceofimage

=

u

v

↦m=

−15

−30

↦m=2cm

Thus;

The magnification will be 2 cm .

[/tex]

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