What is expand ((20-t)(-t+10))/(400) ?

The solution to expand ((20-t)(-t+10))/(400) is (t^2)/(400)-(3t)/(40)+1/2

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expand ((20-t)(-t+10))/(400)

expand

\frac{( 20 - t ) ( - t + 10 )}{400}

\frac{t ^{2}}{400} - \frac{3 t}{40} + \frac{1}{2}

Solution steps

\frac{( 20 - t ) ( - t + 10 )}{400}

Expand

(20 - t) (- t + 10) : - 30 t + t^{2} + 200

= \frac{- 30 t + t ^{2} + 200}{400}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 30 t + t ^{2} + 200}{400} = - \frac{30 t}{400} + \frac{t ^{2}}{400} + \frac{200}{400}

= - \frac{30 t}{400} + \frac{t ^{2}}{400} + \frac{200}{400}

Cancel

\frac{30 t}{400} : \frac{3 t}{40}

= - \frac{3 t}{40} + \frac{t ^{2}}{400} + \frac{200}{400}

Cancel

\frac{200}{400} : \frac{1}{2}

= - \frac{3 t}{40} + \frac{t ^{2}}{400} + \frac{1}{2}

Rewrite in standard form

= \frac{t ^{2}}{400} - \frac{3 t}{40} + \frac{1}{2}

2 (3 x + 6 y = 33)

\frac{3 x + 1}{( x - 1 ) ( x - 2 )}

6 \cdot (- 3 e^{6 t} cos (3 t) + 9 e^{6 t} sin (3 t))

- 2 (3 x - y + 4)^{2}

\frac{4}{( s ^{2} + s + 4 ) s}

What is expand ((20-t)(-t+10))/(400) ?
The solution to expand ((20-t)(-t+10))/(400) is (t^2)/(400)-(3t)/(40)+1/2