Step-by-step explanation:
55x+6=120
55x=120-6
55x=114
X=114/55
-5 because u subtract them both and they equal it ha
So, to find the solution to this problem, we will we using pretty much the same method we used in your previous question. First, let's find the area of the rectangle. The area of a rectangle is length x width. The length in this problem is 16 and the width is 3, and after multiplying these together, we have found 48 in^2 to be the area of the square. Next, we can find the area of the trapezoid. The area of a trapezoid is ((a+b)/2)h where a is the first base, b is the second base, and h is the height. In this problem, a=16, b=5, and h=10. So, all we have to do is plug these values into the area formula. ((16+5)/2)10 = (21/2)10 = 105. So, the area of the trapezoid is 105 in^2. Now after adding the two areas together (48in^2 and 105in^2), we have found the solution to be 153in^2. I hope this helped! :)
Find the average :8.4+7.5+12.9+10.6 divided by four
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