Respuesta :
notice, the volume of the area to be gravelled is just length*width*height.
the length is 2miles, there are 63360 inches in a mile so (2*63360) inches then.
the width is 35 feet, there are 12 inches in a foot, so (12 * 35) inches then.
the height is just 3 inches.
so the volume is (2*63360)(12*35)(3), or 159667200 inches.
now [tex]\bf \begin{array}{ll} yd&in\\ \text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\ yd&36in\\ (yd)^2&(36in)^2\\ (yd)^3&(36in)^3\\ yd^3&36^3in^3\\ &46656in^3 \end{array}[/tex]
how many yd³ is there in 159667200 in³?
[tex]\bf \begin{array}{ccll} yd^3&in^3\\ \text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\ 1&46656\\ x&159667200 \end{array}\implies \cfrac{1}{x}=\cfrac{46656}{159667200}\implies \cfrac{159667200}{46656}=x \\\\\\ \cfrac{30800}{9}=x\\\\ -------------------------------\\\\[/tex]
[tex]\bf \textit{there are }8~yd^3\textit{ in one load, how many loads in }\cfrac{30800}{9}~yd^3? \\\\\\ \cfrac{\frac{30800}{9}}{8}\implies \cfrac{\frac{30800}{9}}{\frac{8}{1}}\implies \cfrac{30800}{9}\cdot \cfrac{1}{8}\implies \cfrac{30800}{72}\implies \cfrac{3850}{9} \\\\\\ \stackrel{loads}{427\frac{7}{9}}[/tex]
the length is 2miles, there are 63360 inches in a mile so (2*63360) inches then.
the width is 35 feet, there are 12 inches in a foot, so (12 * 35) inches then.
the height is just 3 inches.
so the volume is (2*63360)(12*35)(3), or 159667200 inches.
now [tex]\bf \begin{array}{ll} yd&in\\ \text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\ yd&36in\\ (yd)^2&(36in)^2\\ (yd)^3&(36in)^3\\ yd^3&36^3in^3\\ &46656in^3 \end{array}[/tex]
how many yd³ is there in 159667200 in³?
[tex]\bf \begin{array}{ccll} yd^3&in^3\\ \text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\ 1&46656\\ x&159667200 \end{array}\implies \cfrac{1}{x}=\cfrac{46656}{159667200}\implies \cfrac{159667200}{46656}=x \\\\\\ \cfrac{30800}{9}=x\\\\ -------------------------------\\\\[/tex]
[tex]\bf \textit{there are }8~yd^3\textit{ in one load, how many loads in }\cfrac{30800}{9}~yd^3? \\\\\\ \cfrac{\frac{30800}{9}}{8}\implies \cfrac{\frac{30800}{9}}{\frac{8}{1}}\implies \cfrac{30800}{9}\cdot \cfrac{1}{8}\implies \cfrac{30800}{72}\implies \cfrac{3850}{9} \\\\\\ \stackrel{loads}{427\frac{7}{9}}[/tex]
Less than one load of gravel is needed to cover the road.
What is volume?
The amount of space occupied by any three-dimensional solid is called volume.
Formula for the volume of cuboid:
volume = length × breadth × height
Unit conversion:
Conversion of units means the conversion between different units and measurements of the same quantity done by the process of multiplication or division.
Some relationships
1feet = 0.000189 miles
1inch = 1.578 ×[tex]10^{-5}[/tex] miles
1 mile = 1760 yards
According to the given question
Length of the road = 2miles
Width of the road = 35feet = 35×0.000189miles = 0.006615 miles
Depth of the road = 3 inch = 3 × 1.578 ×[tex]10^{-5}[/tex] = 0.000047 miles
Therefore,
The volume of the road = length × depth ×width
= 2 × 0.006615 ×0.000047
=0.00000062181 cubic miles
= 0.010912 cubic yards
For number of loads of gravel will be needed
= volume of gravel to cover the road/ capacity of truck for one load
= 0.010912/8
= 0.001364 load
Hence, less than one load of gravel is needed to cover the road.
Learn more about volume here:
https://brainly.com/question/13338592
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