K=4h+6hp solve for h
1 answer:
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Answer:
r = -12cos(θ)
Step-by-step explanation:
The usual translation can be used:
- x = r·cos(θ)
- y = r·sin(θ)
Putting these relationships into the formula, we have ...
(r·cos(θ) +6)² +(r·sin(θ))² = 36
r²·cos(θ)² +12r·cos(θ) +36 +r²·sin(θ)² = 36
r² +12r·cos(θ) = 0 . . . . subtract 36, use the trig identity cos²+sin²=1
r(r +12cos(θ)) = 0
This has two solutions for r:
r = 0 . . . . . . . . a point at the origin
r = -12cos(θ) . . . the circle of interest
Okay, this is my proof. I'm not exactly sure if this is a viable proof, but I think it works.
Hence, from x > 0, it is always increasing (gradient > 0)
y = lnx crosses the x-axis only once, so there is only one root.
Since x cannot be less than zero, as well as a monotonic increasing function for x > 0, and the fact that it crosses the x-axis once, then as x approaches 0 from the positive side, f(x) has to be approaching negative infinity.
Hence, from x > 0, it is always increasing (gradient > 0)
y = lnx crosses the x-axis only once, so there is only one root.
Since x cannot be less than zero, as well as a monotonic increasing function for x > 0, and the fact that it crosses the x-axis once, then as x approaches 0 from the positive side, f(x) has to be approaching negative infinity.