The answer is <span>he lengths of the bases of the trapezoids are
trapezoid A
b=19cm, and B=35cm, b the short base and B the long base
trapezoid B
</span><span>b=19cm, and B=35cm, b the short base and B the long base
</span>
the height is h=15cm
the surface of A is A= hx(b+B) /2
so the surface is A= 15(19+35)/2=405cm²
we know that suface A =surfaceB, so Surface B= 405cm²
finally the total surface painted is 2 x 405cm²= 810cm²
the perimeter of the trapezoid A is PA=17+19+35+17=88cm
<span>the perimeter of the trapezoid B is PB=17+19+35+17=88cm
</span><span> one piece of sandpaper will not be enough to sand the edges of one trapezoid</span>
Each edge of the inner square has a length of √[(x/2)²+(x/2)²]=√2x²/4=(x√2)/2
Then, the area is 2x²/4=x²/2 ☺☺☺☺
Plug them in
(5)(2)[(4(3)-2(2)]
Now you solve
10(12-4)
Distribute
120-40
Solve
A. 80
A
The domain is -∞ < x < ∞
B
The range is -∞ < x ≤ 3
C
The graph is increasing from -∞ < y < 3
D
The graph is decreasing from 3 > y > -∞
E
The local maximum is at ( - 2, 3 )
F
There are no local minimums