Plsss hurry i need help brainliest involed
1 answer:
You might be interested in
Answer:
The shadow is decreasing at the rate of 3.55 inch/min
Step-by-step explanation:
The height of the building = 60ft
The shadow of the building on the level ground is 25ft long
Ѳ is increasing at the rate of 0.24°/min
Using SOHCAHTOA,
Tan Ѳ = opposite/ adjacent
= height of the building / length of the shadow
Tan Ѳ = h/x
X= h/tan Ѳ
Recall that tan Ѳ = sin Ѳ/cos Ѳ
X= h/x (sin Ѳ/cos Ѳ)
Differentiate with respect to t
dx/dt = (-h/sin²Ѳ)dѲ/dt
When x= 25ft
tanѲ = h/x
= 60/25
Ѳ= tan^-1(60/25)
= 67.38°
dѲ/dt= 0.24°/min
Convert the height in ft to inches
1 ft = 12 inches
Therefore, 60ft = 60*12
= 720 inches
Convert degree/min to radian/min
1°= 0.0175radian
Therefore, 0.24° = 0.24 * 0.0175
= 0.0042 radian/min
Recall that
dx/dt = (-h/sin²Ѳ)dѲ/dt
= (-720/sin²(67.38))*0.0042
= (-720/0.8521)*0.0042
-3.55 inch/min
Therefore, the rate at which the length of the shadow of the building decreases is 3.55 inches/min
Answer:
see explanation
Step-by-step explanation:
Parallel lines have equal slopes.
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
L₁ is y = 5x + 1 ← in slope- intercept form
with slope m = 5
L₂ is 2y - 10x + 3 = 0 ( subtract - 10x + 3 from both sides )
2y = 10x - 3 ( divide all terms by 2 )
y = 5x - ← in slope- intercept form
with slope m = 5
Since L₁ and L₂ have equal slopes then they are parallel lines