True or false
21%
17%
39%
Equation B: 4y = 2 − 4z
Step 1: −4(y) = −4(4 − 2z) [Equation A is multiplied by −4.]
4y = 2 − 4z [Equation B]
Step 2: −4y = 4 − 2z [Equation A in Step 1 is simplified.]
4y = 2 − 4z [Equation B]
Step 3: 0 = 6 − 6z [Equations in Step 2 are added.]
Step 4: 6z = 6
Step 5: z = 1
In which step did the student first make an error?
Step 1
Step 3
Step 4
Step 2
1's answer: 3rd one
2nd one answer: -25
3rd one answer: 1st one
4th one answer: --3
5th one answer: 3rd one.
Hope this helps.
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%.