Until now, given a function <span>f(x)</span>, you would plug a number or another variable in for x. You could even get fancy and plug in an entire expression for x. For example, given <span>f(x) = 2x + 3</span>, you could find <span>f(y2 – 1)</span> by plugging<span> y2 – 1</span> in for x to get <span>f(y2 – 1) = 2(y2 – 1) + 3 = 2y2 – 2 + 3 = 2y2 + 1</span>.
In function composition, you're plugging entire functions in for the x. In other words, you're always getting "fancy". But let's start simple. Instead of dealing with functions as formulas, let's deal with functions as sets of<span> (x, y)</span><span> points </span>
<span>Hope this awnsers your question</span>
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Answer:
1262.0288
Step-by-step explanation:
2×3.14×200.96
RemarkIf you don't start exactly the right way, you can get into all kinds of trouble. This is just one of those cases. I think the best way to start is to divide both terms by x^(1/2)
Step OneDivide both terms in the numerator by x^(1/2)
y= 6x^(1/2) + 3x^(5/2 - 1/2)
y =6x^(1/2) + 3x^(4/2)
y = 6x^(1/2) + 3x^2 Now differentiate that. It should be much easier.
Step TwoDifferentiate the y in the last step.
y' = 6(1/2) x^(- 1/2) + 3*2 x^(2 - 1)
y' = 3x^(-1/2) + 6x I wonder if there's anything else you can do to this. If there is, I don't see it.
I suppose this is possible.
y' = 3/x^(1/2) + 6x
y' =

Frankly I like the first answer better, but you have a choice of both.
Answer:
D
Step-by-step explanation:
no explination
<span>C) Right
https://www.desmos.com/calculator/vf2qksnpwe
</span>right triangle
<span>Proof: You can apply the Pythagorean theorem.</span>