Answer:
Step-by-step explanation:
Polynomial f(x) has the following conditions: zeros of -4 (multiplicity 3), 1 (multiplicity 1), and with f(0) = 320.
The first part zeros of -4 means (x+4) and multiplicity 3 means (x+4)^3.
The second part zeros of 1 means (x-1) and multiplicity 1 means (x-1).
The third part f(0) = 320 means substituting x=0 into (x+4)^3*(x-1)*k =320
(0+4)^3*(0-1)*k = 320
-64k = 320
k = -5
Combining all three conditions, f(x)
= -5(x+4)^3*(x-1)
= -5(x^3 + 3*4*x^2 + 3*4*4*x + 4^3)(x-1)
= -5(x^4 + 12x^3 + 48x^2 + 64x - x^3 - 12x^2 - 48x - 64)
= -5(x^4 + 11x^3 + 36x^2 + 16x -64)
= -5x^3 -55x^3 - 180x^2 - 80x + 320
Answer:
Increasing
−5<x<−1
−1<x<1
Decreasing
4<x<7
−8<x<−5
Neither Increasing nor Decreasing
1<x<4
Step-by-step explanation:
Hello from MrBillDoesMath!
Answer:
See Discussion below
Discussion:
(sinq + cosq)^2 = => (a +b)^2 = a^2 + 2ab + b^2
(sinq)^2 + (cosq)^2 + 2 sinq* cosq => as (sinx)^2 + (cosx)^2 = 1
1 + 2 sinq*cosq (*)
Setting a = b = q in the trig identity:
sin(a+b) = sina*cosb + cosa*sinb
sin(2q) = (**)
sinq*cosq + cosq*sinq => as both terms are identical
2 sinq*cosq
Combining (*) and (**)
(sinq + cosq)^2 = 1 + 2sinq*cosq => (**) 2sinq*cosq = sqin(2q)
= 1 + sin(2q)
Hence
(sinq + cosq)^2 = 1 + sin(2q) => subtracting 1 from both sides
(sinq + cosq)^2 - 1 = sin(2q)
The last statement is what we are trying to prove.
Thank you,
MrB
Answer:
n= 26, so the answer would be 'B'.
Step-by-step explanation:
n-8=18
n=18+8
n=26
Answer:
6 miles
Step-by-step explanation:
x = miles driven
4.00 + 2.75 * x = 20.50
2.75x = 20.50 - 4.00 = 16.50
so
x = 16.50/2.75 = 6 miles
