Answer:
am really and surely 34ca6
keeping in mind that parallel lines have the same exact slope, hmmm what's the slope of 4x - 2y = 3 anyway? well, let's put it in slope-intercept form firstly.

a)
so we're looking for a line whose slope is 2 and passes through 2,1.

b)
![\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{2\implies \cfrac{2}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{2}}\qquad \stackrel{negative~reciprocal}{-\cfrac{1}{2}}} \\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%20%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bperpendicular%20lines%20have%20%5Cunderline%7Bnegative%20reciprocal%7D%20slopes%7D%7D%0A%7B%5Cstackrel%7Bslope%7D%7B2%5Cimplies%20%5Ccfrac%7B2%7D%7B1%7D%7D%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cstackrel%7Breciprocal%7D%7B%5Ccfrac%7B1%7D%7B2%7D%7D%5Cqquad%20%5Cstackrel%7Bnegative~reciprocal%7D%7B-%5Ccfrac%7B1%7D%7B2%7D%7D%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%20)

Answer:
C im pretty sure
Step-by-step explanation:
Because the square taken out has a 4x4 area which is 16 and the area of the whole square used to be 14x14 which is 196 and if u subtract 16 from that you get 180
To solve the question we shall use the formula for the range given by:
Horizontal range, R=[v²sin 2θ]/g
plugging in our values we get:
500=[160²×sin 2θ]/10
5000=160²×sin 2θ
0.1953=sin 2θ
thus:
arcsin 0.1953=2θ
11.263=2θ
hence:
θ=5.6315°~5.63