Check the picture below.
you can pretty much just count off the grid the units for JK and MI.
now, let's check how long are KI and JM
![\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) K&({{ -4}}\quad ,&{{ 4}})\quad % (c,d) I&({{ -2}}\quad ,&{{ 3}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ KI=\sqrt{[-2-(-4)]^2+[3-4]^2}\implies KI=\sqrt{(-2+4)^2+(3-4)^2} \\\\\\ KI=\sqrt{2^2+(-1)^2}\implies KI=\sqrt{4+1}\implies \boxed{KI=\sqrt{5}}\\\\ -------------------------------](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Blllll%7D%0A%26x_1%26y_1%26x_2%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0AK%26%28%7B%7B%20-4%7D%7D%5Cquad%20%2C%26%7B%7B%204%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0AI%26%28%7B%7B%20-2%7D%7D%5Cquad%20%2C%26%7B%7B%203%7D%7D%29%0A%5Cend%7Barray%7D%5Cqquad%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%7B%7B%20x_2%7D%7D-%7B%7B%20x_1%7D%7D%29%5E2%20%2B%20%28%7B%7B%20y_2%7D%7D-%7B%7B%20y_1%7D%7D%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AKI%3D%5Csqrt%7B%5B-2-%28-4%29%5D%5E2%2B%5B3-4%5D%5E2%7D%5Cimplies%20KI%3D%5Csqrt%7B%28-2%2B4%29%5E2%2B%283-4%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AKI%3D%5Csqrt%7B2%5E2%2B%28-1%29%5E2%7D%5Cimplies%20KI%3D%5Csqrt%7B4%2B1%7D%5Cimplies%20%5Cboxed%7BKI%3D%5Csqrt%7B5%7D%7D%5C%5C%5C%5C%0A-------------------------------)
![\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) J&({{ -7}}\quad ,&{{ 4}})\quad % (c,d) M&({{ -8}}\quad ,&{{ 3}}) \end{array}\qquad % distance value \\\\\\ JM=\sqrt{[-8-(-7)]^2+[3-4]^2}\implies JM=\sqrt{(-8+7)^2+(3-4)^2} \\\\\\ JM=\sqrt{(-1)^2+(-1)^2}\implies \boxed{JM=\sqrt{2}}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Blllll%7D%0A%26x_1%26y_1%26x_2%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0AJ%26%28%7B%7B%20-7%7D%7D%5Cquad%20%2C%26%7B%7B%204%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0AM%26%28%7B%7B%20-8%7D%7D%5Cquad%20%2C%26%7B%7B%203%7D%7D%29%0A%5Cend%7Barray%7D%5Cqquad%20%0A%25%20%20distance%20value%0A%5C%5C%5C%5C%5C%5C%0AJM%3D%5Csqrt%7B%5B-8-%28-7%29%5D%5E2%2B%5B3-4%5D%5E2%7D%5Cimplies%20JM%3D%5Csqrt%7B%28-8%2B7%29%5E2%2B%283-4%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AJM%3D%5Csqrt%7B%28-1%29%5E2%2B%28-1%29%5E2%7D%5Cimplies%20%5Cboxed%7BJM%3D%5Csqrt%7B2%7D%7D)
so, add all sides, and that's the perimeter of the trapezoid.
(1 lb 6 oz) + (1 lb 4 oz) + (12 oz) = 2 lb 22 oz.
16 oz = 1 lb, so the quantity (1 lb - 16 oz) evaluates to zero. This lets us rewrite the sum as (2 lb + 22 oz) +(1 lb - 16oz) =
3 lb 6 oz
-9(3d-2)=-29
-27d+18=-29
-27d=-29-18
-27d=-47
d=-47/-27
d=47/27
Answer: 
<u>Step-by-step explanation:</u>
Minimum: (0, -10)
Maximum: (2, -4)
y = A sin (Bx - C) + D
- Amplitude (A) = (Max - Min)/2
- Period = 2π/B → B = 2π/Period
- Phase Shift = C/B → C = B × Phase Shift
- Midline (D) = (Max + Min)/2
Sin usually starts at (0, 0). For this graph, the midline touches 0 when x = 1 so the Phase Shift = 1.
The area of the central rectangle = 5 * 3 = 15 ins^2
area of each triangle = 1/2*3*3 = 4.6 ins^2
Total area = 2*4.5 + 15 = 24 ins^2 (answer)
