-45, 6.64, 44, 467
Least Greatest
Answer:
f(x) = 3x(x-1)(x-3)
Step-by-step explanation:
f(x) = 3x³-12x²+9x = 3x(x²-4x+3) =
= 3x(x²-x-3x+3) = 3x[x(x-1)-3(x-1)] =
= 3x[(x-1)(x-3)] = 3x(x-1)(x-3)
9514 1404 393
Answer:
x = 6
Step-by-step explanation:
The sum of segments is used:
EF +FG = EG
3x +(5x +16) = 11x -2 . . . . substitute given expressions
8x +18 = 11x . . . . . . . . add 2
18 = 3x . . . . . . . . subtract 8x
6 = x . . . . . . divide by 3
The value of the variable is 6.
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<em>Check</em>
The segment lengths are ...
EF = 3x = 3·6 = 18
FG = 5x+16 = 5·6 +16 = 46
EG = 11x -2 = 11·6 -2 = 64 = 18+46 . . . answer is correct
There would be 14 female and 7 male, so the total number of students is 21.
Answer:
85.5 minutes
Step-by-step explanation:
The amount of an element that will remain after time t can be expressed as a function of initial amount N0, time t, and half life th as;
Nt = N0 × e^(-λt)
Where;
Decay constant λ = ln(2)/th, substituting into the equation;
Nt = N0 × e^(-ln(2)t/th)
We need to make t the subject of formula;
Nt/N0 = e^(-ln(2)t/th)
ln(Nt/N0) = -ln(2)t/th
t = ln(Nt/N0) ÷ -ln(2)/th
Given;
Initial amount N0 = 760g
Final amount Nt = 11 g
Half life th = 14 minutes
the nearest tenth of a minute, would it take the element to decay to 11 grams can be derived using the formula;
t = ln(Nt/N0) ÷ -ln(2)/th
Substituting the given values;
t = ln(11/760) ÷ -ln(2)/14
t = 85.5 minutes
