For this problem,we use the Fundamental Counting Principle. You know that there are 7 digits in a number. In this principle, you have to multiply the possible numbers for every digit. If the first number cannot be zero, then there are 9 possible numbers. So, the value for the first digit is 9. The second digit could be any number but less of 1 because it was used in the 1st digit. So, that would be 10 - 1 = 9. The third digit must be the value in the second digit less than 1. That would be 9 - 1 = 8. And so on and so forth. The solution would be:
9×9×8×7×6×5×4 = 544,320 7-digit numbers
round to the nearest
hundredth of milligram : 4.27
tenth of milligram: 4.3
milligram : 4
Answer:
a) shown
b) h = [sqrt(17) - (5/2)t]²
c) t = 2sqrt(17)/5 seconds
Step-by-step explanation:
V = pi × r² × h
V = pi × 5² × h
V = 25pi × h
a) dV/dt = dV/dh × dh/dt
-5pi × sqrt(h) = 25pi × dh/dt
dh/dt = -sqrt(h)/5
b) 1/sqrt(h) .dh = -5. dt
2sqrt(h) = -5t + c
t = 0, h = 17
2sqrt(17) = 0 + c
c = 2sqrt(17)
2sqrt(h) = -5t + 2sqrt(17)
sqrt(h) = [2sqrt(17) - 5t] ÷ 2
sqrt(h) = sqrt(17) - (5/2)t
Square both sides
h = [sqrt(17) - (5/2)t]²
c) empty: h = 0
0 = [sqrt(17) - (5/2)t]²
sqrt(17) - (5/2)t = 0
(5/2)t = sqrt(17)
t = 2sqrt(17)/5
t = 1.64924225 seconds
sqrt: square root
Answer:
(0,2.5)
Step-by-step explanation:
