What is 1250-0.5x^2+(100(2500-x))/(0.02x+1) ?

The solution to 1250-0.5x^2+(100(2500-x))/(0.02x+1) is (-0.01x^3-0.5x^2-75x+251250)/(0.02x+1)

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simplify 1250-0.5x^2+(100(2500-x))/(0.02x+1)

simplify

1250 - 0.5 x^{2} + \frac{100 ( 2500 - x )}{0.02 x + 1}

\frac{- 0.01 x ^{3} - 0.5 x ^{2} - 75 x + 251250}{0.02 x + 1}

Solution steps

1250 - 0.5 x^{2} + \frac{100 ( 2500 - x )}{0.02 x + 1}

Convert

1250

to fraction

: \frac{1250 ( 0.02 x + 1 )}{0.02 x + 1}

Convert

0.5 x^{2}

to fraction

: \frac{0.5 x ^{2} ( 0.02 x + 1 )}{0.02 x + 1}

= \frac{1250 ( 0.02 x + 1 )}{0.02 x + 1} - \frac{0.5 x ^{2} ( 0.02 x + 1 )}{0.02 x + 1} + \frac{100 ( 2500 - x )}{0.02 x + 1}

Apply the fraction rule:

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

= \frac{1250 ( 0.02 x + 1 ) - 0.5 x ^{2} ( 0.02 x + 1 ) + 100 ( 2500 - x )}{0.02 x + 1}

Simplify

1250 (0.02 x + 1) - 0.5 x^{2} (0.02 x + 1) + 100 (2500 - x) : - 0.01 x^{3} - 0.5 x^{2} - 75 x + 251250

= \frac{- 0.01 x ^{3} - 0.5 x ^{2} - 75 x + 251250}{0.02 x + 1}

(x^{2} - 49) (2 + x - 3)

sin (x) (3 sin^{2} (x) + 4 cos^{2} (x))

(\frac{1.5 k}{k + 1}) + (\frac{1.5}{( k + 1 ) ( k + 2 )})

(6 x - 11 + 4)^{2}

arcsin (- \frac{4 x}{( 1 + x ^{2} ) ^{2}})

What is 1250-0.5x^2+(100(2500-x))/(0.02x+1) ?
The solution to 1250-0.5x^2+(100(2500-x))/(0.02x+1) is (-0.01x^3-0.5x^2-75x+251250)/(0.02x+1)