Answer:
afafasfafd shfgjsuhd shdfhsdfghn isndfhusudhfuh nsu
Step-by-step explanation:
dsfshdfh jishudfnhuj hudsnfhjnshjdnf nsjndfjnsdhjnf jhsdnjfhnsdhjnf usndfhsndhjfnshjn jshbndfjhsednfjhn ddhn hshhdfhdsh hhjis ndfjnsd ndfsjhnfh jinse hdfj sihdfhisnhnsd nsdfnshuidfhisdfhjsihjdfnihjsdhfsdj hdhshi dhsdhfsdh dhujishisdfuyhsdhujnghjigihjsdghnsdih huiedsnfghjixhui dyun9y un9sguyhsdyhugnsdghjigsngnih hunxcvhjio nidjnxcgyhui
Answer:
AREA OF SHADED REGION=AREA OF LARGE PARALLELOGRAM-AREA OF SMALL PARALLELOGRAM
=8*12-(6*4)
=96-24
=72unit^2
Answer:
2,789ft
Step-by-step explanation:
The set up will give a right angled triangle.
The altitude will be the opposite side = 1600 ft
The angle of depression = 35°
approximate direct distance to the stadium is the hypotenuse of the trisngle= x
Using the SOH trig identity to get the distance
sin 35 = opp/hyp
sin 35 = 1600/x
x = 1600/sin35
x = 1600/0.5736
x = 2,789.4ft
Hence mai’s approximate direct distance to the stadium is 2,789ft
We know the y intercept is -6 and we know the slope is positive 4 (rise over run) so the equation is y=4x-6 if we plug in a shaded point (I would choose 0,0 for convenience reasons) 0=-6 since -6 is less than 0, the expression would be y≥4x-6
