Answer:
a, b
Step-by-step explanation:
Answer:
Part 1: 359,007 ft³
Part 2: 216 times smaller
Part 3: 21600%
Step-by-step explanation:
Part 1:
The parameters for the tank are;
The radius of the tank = 70 feet
The volume of a sphere = 4/3·π·r³
Therefore, the volume of a quarter sphere = 1/4×The volume of a sphere
The volume of a quarter sphere = 1/4×4/3·π·r³ = π·r³/3
Plugging in the value for the radius gives
Volume = π×70³/3 = 114,333.33×3.14 = 359,006.7≈ 359,007 ft³.
Part 2:
The dimension of the scale model = 1/6 × Actual dimension
Therefore, we have the radius of the sphere of the scale model = 1/6 × 70
Which gives;
The radius of the sphere of the scale model = 35/3 = 11.67 feet
The volume of the scale model = π·r³/3 = (3.14×11.67³)/3 = 1662.07 ≈ 1662 ft³
The number of times smaller the scale model is than the actual volume = (Actual volume)/(Scale model) = (359,007 ft³)/(1662 ft³) = 216 times
The number of times smaller the scale model is than the actual volume = 216 times = (1/Scale of model)³ = (1/(1/6))³ = 6³.
Part 3:
The percentage of the mock-up, x, to the volume of the actual tank is given as follows
x/100 × 1662 = 359,007
∴ x = 216 × 100 = 21600%
The percentage of the mock-up, to the volume of the actual tank is 21600%.
Answer:
Part 1: 359,007 ft³
Part 2: 216 times smaller
Part 3: 21600%
Step-by-step explanation:
Part 1:
The parameters for the tank are;
The radius of the tank = 70 feet
The volume of a sphere = 4/3·π·r³
Therefore, the volume of a quarter sphere = 1/4×The volume of a sphere
The volume of a quarter sphere = 1/4×4/3·π·r³ = π·r³/3
Plugging in the value for the radius gives
Volume = π×70³/3 = 114,333.33×3.14 = 359,006.7≈ 359,007 ft³.
Part 2:
The dimension of the scale model = 1/6 × Actual dimension
Therefore, we have the radius of the sphere of the scale model = 1/6 × 70
Which gives;
The radius of the sphere of the scale model = 35/3 = 11.67 feet
The volume of the scale model = π·r³/3 = (3.14×11.67³)/3 = 1662.07 ≈ 1662 ft³
The number of times smaller the scale model is than the actual volume = (Actual volume)/(Scale model) = (359,007 ft³)/(1662 ft³) = 216 times
The number of times smaller the scale model is than the actual volume = 216 times = (1/Scale of model)³ = (1/(1/6))³ = 6³.
Part 3:
The percentage of the mock-up, x, to the volume of the actual tank is given as follows
x/100 × 1662 = 359,007
∴ x = 216 × 100 = 21600%
The percentage of the mock-up, to the volume of the actual tank is 21600%.
B. The graph would shift down 5 units.
C. The graph would shift up 5 units.
D. The graph would shift 5 units to the left.
Consider the function y=x^2+5x-2 . What would happen to the graph if (x + 5) was substituted in place of the x?
A. The graph would shift 5 units to the right.
B. The graph would shift down 5 units.
C. The graph would shift up 5 units.
D. The graph would shift 5 units to the left.
Option: D is the correct answer.
D. The graph would shift 5 units to the left.
We know that the transformation of a function f(x) to f(x+a) is a shift of the function either to the left or to the right by a units depending on the sign of a i.e. if a>0 then the shift is a units to the left
and if a<0 then the shift is a units to the right.
Here we have:
is converted to:
Here we have a=5>0
Hence, the shift is 5 units to the left.
of B?
Answer:
-3,8
Step-by-step explanation: