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Art [367]
3 years ago
5

Which method do you prefer to use to find sums- count by tens and ones, use compatible numbers, or use friendly numbers and adju

st? Explain why.
Mathematics
2 answers:
Vadim26 [7]3 years ago
8 0
I prefer to use compatible numbers to find sums, as I find this tends to be a faster and more efficient method through which to get to the answer.
Usimov [2.4K]3 years ago
7 0

I prefer to use compatible numbers because by using this method it is easier to make a sum mentally. This is true because compatible numbers are close in value to the actual numbers. For a better understanding, let's take an example:


Suppose you have two numbers, namely 640 and 40. These two numbers are compatible for division because:


64 ÷ 4 = 16


So, we have used mental arithmetic to solve a more complex problem.

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y = \stackrel{\stackrel{m}{\downarrow }}{-\cfrac{2}{3}}x+6\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}

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(\stackrel{x_1}{-3}~,~\stackrel{y_1}{8})\hspace{10em} \stackrel{slope}{m} ~=~ -\cfrac{2}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{8}=\stackrel{m}{-\cfrac{2}{3}}(x-\stackrel{x_1}{(-3)})\implies y-8=-\cfrac{2}{3}(x+3)

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