Answer:
ithink its a or c
Step-by-step explanation:
not sure but dang what grade u in??
Let h represent the height of the trapezoid, the perpendicular distance between AB and DC. Then the area of the trapezoid is
Area = (1/2)(AB + DC)·h
We are given a relationship between AB and DC, so we can write
Area = (1/2)(AB + AB/4)·h = (5/8)AB·h
The given dimensions let us determine the area of ∆BCE to be
Area ∆BCE = (1/2)(5 cm)(12 cm) = 30 cm²
The total area of the trapezoid is also the sum of the areas ...
Area = Area ∆BCE + Area ∆ABE + Area ∆DCE
Since AE = 1/3(AD), the perpendicular distance from E to AB will be h/3. The areas of the two smaller triangles can be computed as
Area ∆ABE = (1/2)(AB)·h/3 = (1/6)AB·h
Area ∆DCE = (1/2)(DC)·(2/3)h = (1/2)(AB/4)·(2/3)h = (1/12)AB·h
Putting all of the above into the equation for the total area of the trapezoid, we have
Area = (5/8)AB·h = 30 cm² + (1/6)AB·h + (1/12)AB·h
(5/8 -1/6 -1/12)AB·h = 30 cm²
AB·h = (30 cm²)/(3/8) = 80 cm²
Then the area of the trapezoid is
Area = (5/8)AB·h = (5/8)·80 cm² = 50 cm²
Answer:
the oeange will be there for u
Step-by-step explanation:
1 then 4 then 3 then 6 then 2
Answer:
b
Step-by-step explanation:
8x +4(x -1)
<em>Expand by multiplying 4 into each term inside the bracket:</em>
= 8x +4(x) +4(-1)
= 8x +4x -4
<em>Simplify</em>
= 12x -4
Thus, the answer is b.
Working with fractions, you'll need to find the least common denominator (LCD). In this case, the LCD is 20.
Rewriting the four given numbers to the LCD, you would get:
10/20, 15/20, 4/20, and 14/20.
Now, you just organize it from least to greatest as:
4/20 < 10/20 < 14/20 < 15/20.
In other words, it is:
1/5 < 1/2 < 7/10 < 3/4.
