This I my explanation for this question if it is helpful your welcome
Answer:
68
Step-by-step explanation:
Draw a line from(-4,3) to (4,3).
a square and triangle is shown.
find the area of the square and the triangle
FORMULA
<em>square-length*width</em>
<em>7*8=56</em>
<em>triangle-(hight*base)/2</em>
<em>(8*3)/2=12</em>
56+12=68
Simplify 2/7b to 2b/7
-1 × (2b/7 + 9) - 8
Simplify 1 × (2b/7 + 9) to (2b/7 + 9)
-(2b/7 + 9) - 8
Simplify brackets
-2b/7 - 9 - 8
Collect like terms
-2b/7 + (-9 - 8)
Simplify
<u>-2b/7 - 17</u>
Since ABCD is a parallelogram, line AB and line DC are parallel and has the same value.
To solve this, equate line AB to be equal with line DC.
So,
Line AB = Line DC
(9x-14)in = (3x +4)in
Next group like terms to get the value of x
9x in-3x in = 4in+14in

=

x = 3in
Since, we now have the value of x, substitute it to line DC’s equation.
DC=(3x+4)in
DC=(3(3) +4)in
DC=(9 +4) in
DC= 13 in
To check if the value is really correct, substitute X to AB
AB=(9x -14)in
AB=(9(3)-14)in
AB=(27-14)in
AB=13 in
Answer:
A cross-section parallel to the base is a rectangle measuring 15 inches by 8 inches.
A cross-section perpendicular to the base through the midpoints of the 8-inch sides is a rectangle measuring 6 inches by 15 inches.
A cross-section not parallel to the base that passes through opposite 6-inch edges is a rectangle measuring 6 inches by greater than 15 inches.
Step-by-step explanation:
the cross sections that are parallel and perpendicular will have the same measurements as the non-intersected sides. the last one will be a diagonal so the intersected edge is 6 and it creates a right triangle so it must be larger than 15 inches.