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enot [183]
2 years ago
5

Make n the subject of the formula

Mathematics
1 answer:
kogti [31]2 years ago
3 0

Answer:

\mathsf{ \frac{m + 21}{5} = n }

Step-by-step explanation:

To make "n" the subject of the formula, rearrange the formula so it begins with " n = "

To isolate the variable "n", you need to inverse the other terms on that side of the equation where "n" really is.

=> m = 5n - 21

  • <em>there's</em><em> </em><em>a</em><em> </em><em>"</em><em>-21</em><em>"</em><em> </em><em>next</em><em> </em><em>to</em><em> </em><em>"</em><em>5n</em><em>"</em><em>,</em><em> </em><em>inverse</em><em> </em><em>of</em><em> </em><em>subtraction</em><em> </em><em>is</em><em> </em><em>addition</em><em>,</em><em> </em><em>so</em><em> </em><em>add</em><em> </em><em>21</em><em> </em><em>on</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>equation</em><em>.</em><em> </em>

=> m + 21 = 5n <u>-</u> <u>21</u> + <u>21</u>

=> m+ 21 = 5n

  • <em>There's</em><em> </em><em>also</em><em> </em><em>a</em><em> </em><em>5</em><em> </em><em>attached</em><em> </em><em>to</em><em> </em><em>n</em><em>,</em><em> </em><em>that</em><em> </em><em>means</em><em> </em><em>the</em><em> </em><em>multiplication</em><em> </em><em>of</em><em> </em><em>n</em><em> </em><em>with</em><em> </em><em>5</em><em>,</em><em> </em><em>the</em><em> </em><em>inverse</em><em> </em><em>of</em><em> </em><em>multiplication</em><em> </em><em>is</em><em> </em><em>division</em><em>,</em><em> </em><em>so</em><em> </em><em>dividing</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>equation</em><em> </em><em>by</em><em> </em><em>5</em>

<em>\mathsf{ \frac{m + 21}{5} = n }</em>

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The function f is such that f(x) = 2x/3x+5 The function g is such that g(x) = 3/x+4 Find fg(-5)
White raven [17]

Answer:

f(g(-5)) = \frac{3}{2}

Step-by-step explanation:

Given

f(x) = \frac{2x}{3x+5}

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Required

Find f(g(-5))

First, we calculate f(g(x))

f(x) = \frac{2x}{3x+5}

Substitute g(x) for x

f(g(x)) = \frac{2g(x)}{3g(x) + 5}

Substitute \frac{3}{x+4} for g(x)

f(g(x)) = \frac{2*\frac{3}{x+4}}{3*\frac{3}{x+4} + 5}

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f(g(x)) = \frac{6}{x+4}/ (\frac{9}{x+4} + 5})

Take LCM

f(g(x)) = \frac{6}{x+4}/ \frac{9+5x+20}{x+4}}

f(g(x)) = \frac{6}{x+4}/ \frac{5x+29}{x+4}}

Rewrite as multiplication

f(g(x)) = \frac{6}{x+4}* \frac{x+4}{5x+29}}

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Substitute -5 for x

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f(g(-5)) = \frac{6}{4}

f(g(-5)) = \frac{3}{2}

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