Answer:
57×6 is 342 ,not sure what you mean by the word expression
Answer:
m = 0
Step-by-step explanation:
AB has a slope of y2 - y1 / x2 - x1 so -8-5/3-3 so -13/0 = undefined so it is a vertical line, so any line that is horizontal will be perpendicular to it, so horizontal lines have a zero slope
It would be 3.5 as you start with 56 then take away 16 =40 then another witch is 24 the another 16 Is 8 so you have taken e16 away so far that's 3 pound and because you can take a full 16 off that makes it a half so final awnser is 3.5 56-16=40-16=24-16=8=3 pounds and a half hope I have helped you in any way
<span>Area of EMIN = 5
The shaded quadrilateral EMIN is composed of two triangles MNI and MNE. Since the area of a triangle is 1/2bh where b is the base and h the height, the area of the quadrilateral is
A = 0.5 * MN * MH + 0.5 MN * EF
A = 0.5 MN (MH + EF)
Now the large rectangle GIKE has an area of 12+36+24+48 = 120. We'll call the overall width "w" and the overall height "h", so wh = 120.
(1) Consider that rectangle GIKE has been divided into 2 rectangles by line FJ. The area above the line is 12+36 = 48, and the area below the line is 24+48 = 72. So the ratio of their heights is 2/3. Therefore we'll call the height of the top rectangle 2h/5 and the bottom rectangle 3h/5.
(2) Considering rectangles FGHM and MHIJ, their heights are the same and their areas are in the ratio 12/36=1/3. So we'll consider the width of FGHM to be w/4 and the width of MHIJ to be 3w/4.
(3) Now consider rectangles EFNL and LNJK. They also share the same height, so their ratios of width is 24/48 = 1/2. So their widths will be w/3 and 2w/3 respectively.
(4) This means that the line segment MN will be w/3 - w/4 = 4w/12 - 3w/12 = w/12 (See (2) and (3) above)
Now taking the formula at the very beginning for the area of the quadrilateral EMIN, we have
A = 0.5 MN (MH + EF)
MN = w/12 ; See (4) above
MH = 2h/5 ; See (1) above
EF = 3h/5 ; See (1) above
So
A = 0.5 MN (MH + EF)
A = 0.5 w/12 (2h/5 + 3h/5)
A = 0.5 w/12 (5h/5)
A = 0.5 w/12 h
A = w/24 h
A = wh/24
And since the overall area of the entire rectangle is wh = 120, we get
A = wh/24
A = 120/24
A = 5
As for the actual width and height of the overall rectangle? Don't know and don't care. Any value of w and h will work as long as their product is 120. But regardless of the actual dimensions, the area of quadrilateral EMIN will be 5.</span>