Answer:
The dimensions of the Park is120m2
10(10x−12)=−9(−9x−2)−5
Step 1: Simplify both sides of the equation.
−10(10x−12)=−9(−9x−2)−5
(−10)(10x)+(−10)(−12)=(−9)(−9x)+(−9)(−2)+−5(Distribute)
−100x+120=81x+18+−5
−100x+120=(81x)+(18+−5)(Combine Like Terms)
−100x+120=81x+13
−100x+120=81x+13
Step 2: Subtract 81x from both sides.
−100x+120−81x=81x+13−81x
−181x+120=13
Step 3: Subtract 120 from both sides.
−181x+120−120=13−120
−181x=−107
Step 4: Divide both sides by -181.
−181x
−181
=
−107
−181
x=
107
181
Answer:
x=
107
181
Answer:
The point C is 12.68 km away from the point A on a bearing of S23.23°W.
Step-by-step explanation:
Given that AB is 50 km and BC is 40 km as shown in the figure.
From the figure, the length of x-component of AC = |AB sin 80° - BC cos 20°|
=|50 sin 80° - 40 cos 20°|=11.65 km
The length of y-component of AC = |AB cos 80° - BC sin 20°|
=|50 cos 80° - 40 sin 20°|= 5 km
tan
= 5/11.65
=23.23°
AC=
km
Hence, the point C is 12.68 km away from the point A on a bearing of S23.23°W.
Fourteen hours. The equation is 5x+20=90. Solve for X.