Answer:
62°
Step-by-step explanation:
The angle R inscribes the arc FQ, so using the property of inscribed angles in a circle, we have that:
m∠R = mFQ / 2
The arc FQ is the sum of the arcs FP and PQ, so we have:
mFQ = mFP + mPQ = 11x + 7 + 60 = 11x + 67
Now, with the first equation, we have:
12x + 1 = (11x + 67) / 2
24x + 2 = 11x + 67
13x = 65
x = 5°
So we have that mFP = 11x + 7 = 55 + 7 = 62°
Alright, this one is a little interesting... Let's perform some tests to figure out what is happening:
f(-10) = -(1/(-10)^3) = -(1/-1000) = 1/1000 (positive)
f(-5) = -(1/(-5)^3) = -(1/-125) = 1/125 (positive, bigger than the last one)
f(-1) = -(1/(-1)^3) = -(1/-1) = 1 (positive, bigger than the last one)
f(-0.1) = -(1/(-0.1)^3) = -(1/-0.001) = 1/0.001 = 1000 (positive, bigger than the last one)
f(0) = -(1/0^3) = undefined!
f(0.1) = -(1/(0.1)^3) = -(1/0.001) = -1/0.001 = -1000 (negative)
f(1) = -(1/1^3) = -(1/1) = -1 (negative, but bigger than last one)
It's a little confusing with the undefined part at x = 0. What I can say is this, it is increasing from -10 up to 0, something weird happens at 0 and it resets, and starts increasing from 0 up to 0.1.
I guess A would be the best answer?
Answer: Choice B. k(h(g(f(x))))
For choice B, the functions are k, h, g, f going from left to right.
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Explanation:
We have 4x involved, so we'll need f(x)
This 4x term is inside a cubic, so we'll need g(x) as well.
So far we have
g(x) = x^3
g( f(x) ) = ( f(x) )^3
g( f(x) ) = ( 4x )^3
Then note how we are dividing that result by 2. That's the same as applying the h(x) function
And finally, we subtract 1 from this, but that's the same as using k(x)
This leads to the answer choice B.
To be honest, this notation is a mess considering how many function compositions are going on. It's very easy to get lost. I recommend carefully stepping through the problem and building it up in the way I've done above, or in a similar fashion. The idea is to start from the inside and work your way out. Keep in mind that PEMDAS plays a role.
Is this the actual question
