Answer:
Enter the value of the score in the laboratory 1
L1 = digited value.
Enter the value of the score in laboratory 2
L2 = digited value.
Enter the value of the score in laboratory 3
L3 = digited value.
Enter the value of the test score 1
P1 = digited value.
Enter the value of the test score 2
P2 = digited value.
Laboratory Average = (L1 + L2 + L3) / 3
Print "Laboratory Average"
Test Average = (P1 + P2) / 2
Print "Test Average"
Course qualification = 0.55 * (Laboratory Average) + 0.45 * (Test Average)
print "Course qualification"
Step-by-step explanation:
(1): "c0" was replaced by "c^0".
(2): Dot was discarded near "1.j". 1 more similar replacement(s)
Unauthorized use of the imaginary unit "i" or syntax error in complex arithmetic expression
Answer:
(-6, 0)
Step-by-step explanation:
(x, y) over y=x = (y, x)
so (0, -6) will be (-6, 0).
Answer:
picture ?
Step-by-step explanation:
Answer: 46.90mins
Step-by-step explanation:
The given data:
The diameter of the balloon = 55 feet
The rate of increase of the radius of the balloon when inflated = 1.5 feet/min.
Solution:
dr/dt = 1.5 feet per minute = 1.5 ft/min
V = 4/3·π·r³
The maximum volume of the balloon
= 4/3 × 3.14 × 55³
= 696556.67 ft³
When the volume 2/3 the maximum volume
= 2/3 × 696556.67 ft³
= 464371.11 ft³
The radius, r₂ at the point is
= 4/3·π·r₂³
= 464371.11 ft³
r₂³ = 464371.11 ft³ × 3/4
= 348278.33 ft³
348278.333333
r₂ = ∛(348278.33 ft³) ≈ 70.36 ft
The time for the radius to increase to the above length = Length/(Rate of increase of length of the radius)
The time for the radius to increase to the
above length
Time taken for the radius to increase the length.
= is 70.369 ft/(1.5 ft/min)
= 46.90 minutes
46.90mins is the time taken to inflate the balloon.
