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Wewaii [24]
3 years ago
9

The LCM of 12ab² and 3ab³

Mathematics
1 answer:
lions [1.4K]3 years ago
8 0

Answer:3ab^2 or 12ab^3

i think this is it

Step-by-step explanation:

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Help. I need help with these questions ( see image).<br> Please show workings.
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9514 1404 393

Answer:

  4)  6x

  5)  2x +3

Step-by-step explanation:

We can work both these problems at once by finding an applicable rule.

  \text{For $f(x)=ax^n$}\\\\\lim\limits_{h\to 0}\dfrac{f(x+h)-f(x)}{h}=\lim\limits_{h\to 0}\dfrac{a(x+h)^n-ax^n}{h}\\\\=\lim\limits_{h\to 0}\dfrac{ax^n+anx^{n-1}h+O(h^2)-ax^n}{h}=\boxed{anx^{n-1}}

where O(h²) is the series of terms involving h² and higher powers. When divided by h, each term has h as a multiplier, so the series sums to zero when h approaches zero. Of course, if n < 2, there are no O(h²) terms in the expansion, so that can be ignored.

This can be referred to as the <em>power rule</em>.

Note that for the quadratic f(x) = ax^2 +bx +c, the limit of the sum is the sum of the limits, so this applies to the terms individually:

  lim[h→0](f(x+h)-f(x))/h = 2ax +b

__

4. The gradient of 3x^2 is 3(2)x^(2-1) = 6x.

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If you need to "show work" for these problems individually, use the appropriate values for 'a' and 'n' in the above derivation of the power rule.

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