Answer:
Teo is 11 years old and Richard is 26 years old.
Step-by-step explanation:
Let R be the age of Richard and T be the age of Teo.
<u><em>First set up the equations:</em></u>
The combined age of Teo and Richard is 37:
R+T=37
Richard is four years old than twice Teo's age:
R=2T+4
<u><em>Then, solve the equations:</em></u>
Substitute (2) to (1) and solve for T:
(2T+4)+T=37
3T=11
T=11
Solve for R using (2):
R=2(11)+4
R=22+4
R=26
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:
f(x) is translated by 4 units left.
Step-by-step explanation:
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Answer
We have,
Volume of pyramid = 4834 in³
Base area of the pyramid = ?
Assuming height of the pyramid = 20 in
We know that
Volume of pyramid =
l w = 725.1 in²
Hence, area of the base of the pyramid is equal to 725.1 in².