The wall area is the product of the room perimeter and the room height:
A₁ = (2*(12.5 ft + 10.5 ft))*(8.0 ft) = 368 ft²
The window and door area together is
A₂ = 2*((4 ft)*(3 ft)) + (7 ft)*(3 ft) = 45 ft²
The area of one roll of wallpaper is
A₃ = (2.5 ft)*(30 ft) = 75 ft²
Then the number of rolls of wallpaper required will be
1.1*(A₁ - A₂)/A₃ ≈ 4.74
5 rolls of wallpaper should be purchased.
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As a practical matter, not much of the window and door area can be saved. The rolls are 30 inches wide, but the openings are 36 inches wide. Some will likely have to be cut from two strips. The strips will have to be the full length of the wall, and the amount cut likely cannot be used elsewhere. If the window and door area cannot be salvaged, then likely ceiling(5.4) = 6 rolls will be needed (still allowing 10% for matching and waste).
Hey!
91 ÷ 7 = 13
So, we know that Luke paints one portrait every week.
13 × 4 = 52
So, Luke paints 52 portraits in 4 weeks.
Answer:
Reema bought 24 pencils in august.
Step-by-step explanation:
Answer:
A = 6
So
Side A = 6
Side B = 11
Side C = 3
Side D = 13
Explanation:
To solve perimeter of a quadrilateral, you use this formula:
side A + side B + side C + side D = total perimeter.
If the four sides are A, A + 5, A – 3, and 2A + 1 then you need to solve for A.
Write out an algebraic formula with these values:
A + (A + 5) + (A – 3) + (2A + 1) = P
You also know that the total perimeter is 33, so :
A + (A + 5) + (A – 3) + (2A + 1) = 33
You can add all the A values together to combine like terms like this:
5A + 5 – 3 + 1 = 33
Solve fore A:
5A + 3 = 33
5A = 33 – 3
5A = 30
A = 30 ÷ 5
A = 6
Check your answer:
A + (A + 5) + (A – 3) + (2A + 1) = 33
(6) + ((6) + 5) + ((6) – 3) + (2(6) + 1) = 33
(6) + (11) + (3) + (13) = 33
17 + 16 = 33
33 = 33
