(6x5)−13≤−13(\frac{6x}{5}) - \frac{1}{3} \leq -\frac{1}{3}(56x)−31≤−31
7x3-9x4+(-2)x6+3x8
−26m−15mn−35p+70pq÷5
A point D D D is chosen on side BC BC BC of triangle ABC ABC ABC with incentre I I I.
The perpendicular bisector of BC BC BC ...
18abc−19bc+21c÷6c
solve for 7(x+2)=7
Ordenada al origen: 20; abscisa al origen: 40; pendiente: -0,5
x⁴
x3245
