Let: x = the number of games,
y = the number of rides
For 2.50x+y= 15, what's the number of rides the
student could get on if they don't play any games?

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Ответ:

Arealbot

09.08.2021

Mathematics

$\frac{1}{m-4}$

Step-by-step explanation:

$\frac{\frac{4m-5}{m^4 -7m^3 +12m^2}}{\frac{4m-5}{m^3 -3m^2}}$

Factor the equation:

$\frac{\frac{4m-5}{m^2(m^2 -7m+12)}}{\frac{4m-5}{m^2(m-3)}}$

$\frac{\frac{4m-5}{m^2(m-3)(m-4)}}{\frac{4m-5}{m^2(m-3)}}$

Rewrite to suit the format of multiplying two fractions. Remember, dividing two fractions is the same as multiplying the first fraction by the reciprocal of the second. A reciprocal of a fraction is when one switches the place of the numerator and the denominator, that is, the value on top (numerator), and the value on the bottom (denominator).

$\frac{4m-5}{m^2(m-3)(m-4)}*\frac{m^2(m-3)}{4m-5}$

Simplify, take out common terms that are found on both the numerator and denominator

$\frac{4m-5}{m^2(m-3)(m-4)}*\frac{m^2(m-3)}{4m-5}$

$\frac{1}{m-4}$

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Let: x = the number of games, y = the number of rides For 2.50x+y= 15, what's the number of rides the student could get on if they don't play any games?