2cos²x+cos x-1=0
cos x=t
2t²+t-1=0
t=[-1⁺₋√(-1+8)]/2=(-1⁺₋3)/4
We have two possible set solutions:
First set solutions.
t₁=(-1-3)/2=-4/4=-1
cos x=-1 ⇒x=cos⁻¹ (-1)=π +2kπ or 180º+360ºk (k=(...-2,-1,0,1,2...)
Second set solutions:
t₂=(-1+3)/4=2/4=1/2
cos x=1/2 ⇒ x=cos⁻¹ 1/2=π/3+2kπ U 5π/3+ 2kπ or
60º+360ºK U 300º+360ºK (k=...-2,-1,0,1,2,...)
solutions: first set solutions U second set solutions:
Answer in radians : π +2kπ U π/3+2kπ U 5π/3+ 2kπ (k=...-1,0,1,...)
Answer is degrees: 180º+360ºk U 60º+360ºK U 300º+360ºK (k=...-2,-1,0,1,2,...)
Answer:
(d) f(x) = log₆(x)
Step-by-step explanation:
If we use y in place of f(x), we see that ...
x = 6^y
Taking logs of both sides, we get ...
log(x) = y·log(6)
y = log(x)/log(6)
y = log₆(x) . . . . . . . using the change of base formula
f(x) = log₆(x)
_____
Or you can get there more directly using the relation between logs and exponentials:
32g/m = 32*60 = 1920 gallons per hour
1 gallon = 4 quarts
1920 *4 = 7680 quarts per hour
Y = -2/3x + 6....the slope here is -2/3
y - y1 = m(x - x1)
slope(m) = -2/3
(-3,8)....x1 = -3 and y1 = 8
now we sub
y - 8 = -2/3(x - (-3) =
y - 8 = -2/3(x + 3) <===
Bottom left
EXPLANATION: The least value will be the one when you multiply something less than 1. the middle is when you multiply exactly one. and the greatest is when you multiply more than 1 by 564
