(4+3n)⋅(3n−2n)−(n−1)⋅5n=
asked by guest on Mar 30, 2025 at 4:47 am
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(4+3n)(3n−2n)−(n−1)⋅5n=n(3n+4)−5n(n−1)\displaystyle \left(4 + 3 n\right) \left(3 n - 2 n\right) - \left(n - 1\right) \cdot 5 n = n \left(3 n + 4\right) - 5 n \left(n - 1\right) (4+3n)(3n−2n)−(n−1)⋅5n=n(3n+4)−5n(n−1)
(4+3n)(3n−2n)−(n−1)⋅5n=−2n2+9n\left(4 + 3 n\right) \left(3 n - 2 n\right) - \left(n - 1\right) \cdot 5 n = - 2 n^{2} + 9 n(4+3n)(3n−2n)−(n−1)⋅5n=−2n2+9n
(4+3n)(3n−2n)−(n−1)⋅5n=−n(2n−9)\left(4 + 3 n\right) \left(3 n - 2 n\right) - \left(n - 1\right) \cdot 5 n = - n \left(2 n - 9\right)(4+3n)(3n−2n)−(n−1)⋅5n=−n(2n−9)
