I know that: the first term of the sequence is -2, and the fifth term is -1/8
1 answer:
Solution
Here the first term = -2
5th term = -1/8
Now we have to find the common ratio (r)
Fifth term A5 = -1/8
-1/8 =
Here
Plug in in the fifth term and find the value of r.
-1/8 = -2
r^4 = 1/16
Taking 4th root, we get
r = 1/2
Now let's find the second term.
first term *r = -2 *1/2 = -1
2nd term = first term * 1/2 = -1
3rd term = second term * 1/2 = -1 * 1/2 = -1/2
4th term = 3rd term *r = -1/2 *1/2 = -1/4
Thank you :)
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Answer:
m ≥ -√5
Step-by-step explanation:
If g(x) = f(|x|) for x∈i, then g(x) = f(x) for x∈ℝ: x ≥ 0.
f'(x) is 5th-degree, so f(x) is 6th-degree, meaning it is generally U-shaped. Since we're only concerned with x ≥ 0, we want to make sure f'(x) has no real zeros of odd multiplicity such that x > 0. The given factors of f'(x) make it have real zeros at x = -3 and x = -1.
For the last factor, (x² +2mx +5) to have no positive real zeros of odd multiplicity, we must have m ≥ 0 or the discriminant ≤ 0. The discriminant is ...
d = b² -4ac = (2m)² -4(1)(5) = 4m² -20 . . . . . discriminant of the last factor
d ≤ 0 . . . . . . . . . . the condition for no real zeros
4m² -20 ≤ 0
m² -5 ≤ 0 . . . . . . divide by 4
m² ≤ 5 . . . . . . . . .add 5
|m| ≤ √5 . . . . . . . take the square root
This tells us there will be a positive real zero of multiplicity 2 in f'(x) when m = -√5, and there will be no positive real zeros for -√5 < m < 0
There will be no odd-multiplicity positive real zeros in the derivative function f'(x) as long as m ≥ -√5. This means the slope of f(x) is non-negative for x ≥ 0, hence f(|x|) has its only minimum at x=0.
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<em>Additional comment</em>
The multiplicity of the zeros of f'(x) is important because the derivative will only change sign where the multiplicity is odd. When the discriminant of (x²+2mx+5) is zero, the associated positive real zero will have multiplicity 2, hence f'(x) will not change sign there.
Given that the ratio of three trays is A : B : C = 2 : 3 : 4
Let us consider the common factor among the number of trays be "x".
If we need to make a total of 171 trays, then the equation of sum of trays would be :-
Type A + Type B + Type C = 171
2x + 3x + 4x = 171
9x = 171
x = = 19
So we have following number of trays :-
Number of type A trays = 2x = 38 trays.
Number of type B trays = 3x = 57 trays.
Number of type C trays = 4x = 76 trays.
Now if each tray contains 20 cookies, then we would have following arrangements :-
Cookies in type A trays = 38×20 = 760 cookies.
Cookies in type B trays = 57×20 = 1140 cookies.
Cookies in type C trays = 76×20 = 1520 cookies.