The heat in the house is set to keep the minimum and maximum temperatures (in degrees Fahrenheit) according to the equation |x –
2 answers:
You might be interested in
This question is Incomplete because it lacks the diagram showing the containers as well as the required options. Find attached to this answer the appropriate diagram.
Complete Question
Aluminium coated dewars are cylindrical flasks used to safely store and transport liquid nitrogen, dry ice etc. A company manufactures such dewars of different sizes. Shown here are two dewars, both of the same height but of different diameters. How much aluminium will be required to cover Dewar A compared to that required for Dewar B? (Note: The base and the lid are NOT made up of aluminium.)
A) 1/3 times
B) 3 times
C) 9 times
D) The same amount of aluminum will be required.
Answer:
B) 3 times
Step-by-step explanation:
From the question, we are told that we are to find the amount of aluminum to cover Dewar A and B , we are also told that the base and the lid are not covered with aluminum.
From this information, we can determine that we are to find the curved or Lateral surface area of the cylinder because the base and lid are not included
The curved surface area of a cylinder is calculated as 2πrh
We told that both cylinders have the same height. Hence,
For Dewar A
We have Diameter of 30 cm, radius = Diameter ÷ 2 = 30cm ÷ 2 = 15cm.
Height = h
The curved surface area = 2πrh
= 2 × π × 15 ×h
= 30πhcm²
For Dewar B
We have Diameter of 10 cm, radius = Diameter ÷ 2 = 10cm ÷ 2 = 5cm.
Height = h
The curved surface area = 2πrh
= 2 × π × 5 ×h
= 10πhcm²
When we compare the curved surface Dewar A to the curved surface area of Dewar B
Dewar A : Dewar B
30πhcm² : 10πhcm²
3(10πhcm²) : 10πhcm²
From the above comparison, we can see that 3 times more aluminum is required to cover Dewar A compared to Dewar B.
Answer:
The line containing the midpoints of two sides of a triangle is parallel to the third side ⇒ proved down
Step-by-step explanation:
* Lets revise the rules of the midpoint and the slope to prove the
problem
- The slope of a line whose endpoints are (x1 , y1) and (x2 , y2) is
- The mid-point of a line whose endpoints are (x1 , y1) and (x2 , y2) is
* Lets solve the problem
- PQR is a triangle of vertices P (0 , 0) , Q (2a , 0) , R (2b , 2c)
- Lets find the mid-poits of PQ called A
∵ Point P is (x1 , y1) and point Q is (x2 , y2)
∴ x1 = 0 , x2 = 2a and y1 = 0 , y2 = 0
∵ A is the mid-point of PQ
∴
- Lets find the mid-poits of PR which called B
∵ Point P is (x1 , y1) and point R is (x2 , y2)
∴ x1 = 0 , x2 = 2b and y1 = 0 , y2 = 2c
∵ B is the mid-point of PR
∴
- The parallel line have equal slopes, so lets find the slopes of AB and
QR to prove that they have same slopes then they are parallel
# Slope of AB
∵ Point A is (x1 , y1) and point B is (x2 , y2)
∵ Point A = (a , 0) and point B = (b , c)
∴ x1 = a , x2 = b and y1 = 0 and y2 = c
∴ The slope of AB is
# Slope of QR
∵ Point Q is (x1 , y1) and point R is (x2 , y2)
∵ Point Q = (2a , 0) and point R = (2b , 2c)
∴ x1 = 2a , x2 = 2b and y1 = 0 and y2 = 2c
∴ The slope of AB is
∵ The slopes of AB and QR are equal
∴ AB // QR
∵ AB is the line containing the midpoints of PQ and PR of Δ PQR
∵ QR is the third side of the triangle
∴ The line containing the midpoints of two sides of a triangle is parallel
to the third side