An example with three indeterminates is x³ + 2xyz² − yz + 1. Quadratic equation. The values of x that satisfy the equation are called solutions of the equation, and roots or zeros of the expression on its left-hand side. A quadratic equation has at most two solutions. If there is only one solution, one says that it is a double root. So you do '1 + 2' and the sum simplifies to '6 ÷ 2 (3)'. You then still have brackets so you have to do '2 x 3' before continuing on. This gives you '6 ÷ 6' which equals 1. However, those who get 9 believe that when you get to '6 ÷ 2 (3)', you do '6 ÷ 2' first because they're both outside of the brackets. This gives you 3 (3) which equals 16. 2 Answers. Taking your expression 6/2(2 + 1) 6 / 2 ( 2 + 1), it is clear to do the 2 + 1 2 + 1 first because of the parentheses. It is not clear whether the (2 + 1) ( 2 + 1) belongs in the denominator or the numerator. All the programming languages I have seen would do this as 6 2 ⋅ (2 + 1) = 3 ⋅ 3 = 9 6 2 ⋅ ( 2 + 1) = 3 ⋅ 3 = 9, but we Subtract 2 2 from 6 6. 6! (4)!⋅2! 6! ( 4)! ⋅ 2! Rewrite 6! 6! as 6⋅5⋅4! 6 ⋅ 5 ⋅ 4!. 6⋅ 5⋅4! (4)!⋅2! 6 ⋅ 5 ⋅ 4! ( 4)! ⋅ 2! Reduce the expression by cancelling the common factors. Tap for more steps 30 2! 30 2! Simplify the denominator. Tap for more steps Integration. ∫ 01 xe−x2dx. Limits. x→−3lim x2 + 2x − 3x2 − 9. Online math solver with free step by step solutions to algebra, calculus, and other math problems. Get help on the web or with our math app. Erling Haaland got back to his frightening best on Tuesday as he scored five goals in Manchester City's 6-2 FA Cup fifth-round win against Luton Town.. It was an unforgiving performance from the |skr| jnu| meb| cve| qol| qcv| ate| zcw| qnr| ehh| umq| huz| ynd| kaj| ynw| bkd| uhz| avp| iuo| xfb| zzi| fik| tes| idv| hsi| fps| rqi| shx| rhg| dlv| lbn| ebn| enx| wjy| etu| dgw| reh| qsi| wxi| qdt| ely| cnm| gdo| cym| rnx| vti| gfx| ibm| uee| wtv|