I^3=-i
therefor -27i=(3i)^3
so we can get a sum of 2 perfect cubes
a^3+b^3=(a+b)(a^2-ab+b^2)
so
remember that i²=-1
x^3+(3i)^3=(x+3i)(x²-3xi-1)
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:
steps and answers in the pic..hope u can understand the calculations...
Answer:
31 92 123
1 2 3
Step-by-step explanation:
