Area & Surface Area:Question 7 The dimensions for a square pyramid are shown in the figures. What is the lateral surface area of the figure?
Please help

208 think but I haven't done this type of stuff in a long time

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1

answer: cos [   arcsin (3/5) - arccos (1/2) ] =

step-by-step explanation: × = 3/5 = 0.6--> x = 36.87 and x = 180 - 36.87 = 143.13 deg

cos y = 1/2 --> y = +- 60 deg a. cos (x - y) = cos (143.13 - 60) = cos 83,13  

=

0.12

b. cos (36.87 - 60) = cos (-23.23) = 0.92