We have 8 yards of wraping paper. if we use 2 feet for each present how many present can we wrap
2 answers:
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<h3>Answer: 24.5 </h3>
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Explanation:
It might help to draw it out. See the diagram below.
The base is JK which is 7 units long. I'm picking this as the base because this segment is completely horizontal.
Notice how the base JK spans from x = -3 to x = 4, which we can use subtraction and absolute value to get |J-K| = |-3-4| = 7.
The height is always perpendicular to the base. If the base is JK = 7, then the height is the vertical distance from K (or J) to L. This vertical distance is also 7 units. Subtract the y coordinates and apply absolute value. The base and height aren't always the same number.
Now we can find the area of the triangle
area = base*height/2
area = 7*7/2
area = 49/2
area = 24.5 square units
Answer:
a) 38"
b) L = 5.7'
Step-by-step explanation:
a) Given
P = 9,500 lb/ft
Thick = 18"
We assume an allowable bearing pressure of σ = 3000 lb/sf
If Stress = Force / Area
σ = P/A
Solving for area
A = P/σ ⇒ A = 9,500 lb/ft/3000 lb/sf
⇒ A = 3.167 sf, or 3.167 ft wide, per foot of length.
⇒ A = 38" wide
We can see the pic 1 in order to understand the answer.
b) Given
P = 65,000 pounds = 65,000 lb
Thick = 18" = 1.5'
We assume an allowable bearing pressure of σ = 2000 lb/sf
If Stress = Force / Area
σ = P/A
Solving for area
A = P/σ ⇒ A = 65,000 lb/2000 lb/sf
⇒ A = 32.5 sf
then A = L² ⇒ L = √A = √(32.5 sf) = 5.7 ft
Finally L = 5.7'
We can see the pic 2 in order to understand the answer.
