The sum of the lengths of the bases (b1+b2) for the given trapezoid is 10 meters.
<u>Step-by-step explanation:</u>
The given information are,
- The height of the parallelogram = 7 meters
- The area of the parallelogram = 35 square meters.
It is given that the height and area of the trapezoid is same as the height and area of the parallelogram.
<u>To find the sum of the lengths of the bases of trapezoid :</u>
Substitute height, h of the trapezoid as 7 meters and Area of trapezoid as 35 square meters.
Area of the trapezoid = 1/2 × h × (b1+b2)
35 = 1/2 × 7 × (b1+b2)
70/7 = (b1+b2)
(b1+b2) = 10
Therefore, the sum of lengths of the bases of trapezoid is 10 meters.
A foot is 12 inches... so it would be a foot
Answer:
y - 1 = - 4(x + 3)
Step-by-step explanation:
The equation of a line in point- slope form is
y - b = m(x - a)
where m is the slope and (a, b) a point on the line
Here m = - 4 and (a, b) = (- 3, 1), thus
y - 1 = - 4(x - (- 3)), that is
y - 1 = - 4(x + 3)
Https://answers.yahoo.com/question/index?qid=20120624121412AAV4Cjw
Hope that helps!
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Answer:
Step-by-step explanation:
Given that:
where;
the top vertex = (0,0,1) and the base vertices at (0, 0, 0), (1, 0, 0), (0, 1, 0), and (1, 1, 0)
As such , the region of the bounds of the pyramid is: (0 ≤ x ≤ 1-z, 0 ≤ y ≤ 1-z, 0 ≤ z ≤ 1)
![\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \ dz \ ( \dfrac{(1-z)^3}{3} \ y + \dfrac {(1-z)y^3)}{3}] ^{1-x}_{0}](https://tex.z-dn.net/?f=%5Ciiint_W%20%28x%5E2%2By%5E2%29%20%5C%20dx%20%5C%20dy%20%5C%20dz%20%3D%20%5Cint%20%5E1_0%20%20%5C%20dz%20%5C%20%20%28%20%5Cdfrac%7B%281-z%29%5E3%7D%7B3%7D%20%5C%20y%20%2B%20%5Cdfrac%20%7B%281-z%29y%5E3%29%7D%7B3%7D%5D%20%5E%7B1-x%7D_%7B0%7D)
