<h2>Answer: A trapezoid with bases of 6 mm and 14 mm and a height of 8 mm </h2>
The parallelogram in the figure has an area of 
, according to the following formula, which works for all rectangles and parallelograms:
(1)
Where 
is the base and 
is the height
The<u> area of a triangle</u> is given by the following formula:
(2)
So, for option A:
Now, the <u>area of a trapezoid </u>is:
(3)
For option B:
For option C:
>>>>This is the correct option!
For option D:
<h2>Therefore the correct option is C</h2>
The answer is choice A.
We're told that the left and right walls of the cube (LMN and PQR) are parallel planes. Any line contained in one of those planes will not meet another line contained in another plane. With choice A, it's possible to have the front and back walls be non-parallel and still meet the initial conditions. If this is the case, then OS won't be paralle to NR. Similarly, LP won't be parallel to MQ.
The first one is 154
The second one is 12.65^2
I'm not completely sure but this is what I would do.
evaluate <span>(1/ 4)^x - 1 </span>as is. But change the (1 /2)^2x to (2/4)^2x. This way both fractions have the same denominator and in this sense, the same base. The 2/4 base still evaluates into 1/2 so nothing, mathematically, is being broken here.
The answer is D. 9(y + 3) and 27 + 9y because 9(y + 3) when expanded is 9y + 27. I hope this helps!
