Tutor profile: Alec W.
Subject: Basic Math
Round 17.625 to the nearest hundreth.
A contractor is milling and overlaying a 500 foot section of 2-lane road with no breakdown lanes. The lanes are standard width lanes. Given mill and overlay depth of 2", and an asphalt mix density of 140 lb/ft^3, what is the total tonnage of asphalt needed to pave the roadway?
A standard lane is 12 ft wide. This results in a total roadway width of 24'. To find the answer, we must first find the total volume of asphalt to be placed. Length X Width X Depth = Volume 500 FT X 24 FT X (2/12) FT = 2000 ft^3 We then multiply our volume by our density to give us our needed asphalt in pounds. 2000 FT^3 X 140 LB/FT^3 = 280,000 LB We then convert pounds to tons by diving by 2,000. 280,000 LB / 2,000 LB/TON = 140 Tons
2X+3Y=26 4X+Y=32 Solve for X and Y.
1.) Multiply the 1st equation by a multiple of 2. 2(2X+3Y=26) 4X+6Y=52 2.) Use subtraction to eliminate the X variable from the system of equations. 4X+6Y=52 - 4X+Y=32 0X+5Y=20 5Y=20 Y=4 3.) Insert the known Y into either of the original equations to solve for the other variable, X. 2X+3Y=26 2X+(3*4)=26 2X+12=26 2X=14 X=7 Answer: X=7 Y=4
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