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8p - 7q equals 2p - 5 Is this equation linear or nonlinear?
This equation is linear, because neither variable symbol occurs to any power except the first.
Is this equation linear or nonlinear 8p - 7q 2p - 5?
Neither because without an equality sign the given expressions can't be classed as an equation.
Maths algebra what is 8p-p-4p?
8p-p-4p= 7p-4p= 3p
How do you factorise expressions?
If you are factorising expressions you need to look for the highest common factor (hcf);for example15y + 10 = 5(3y+5)hcf = 5(5 x 3y = 15y) + (5 x 2 = 10)(5 x 3y) + (5 x2)Put the hcf at the beginning of the equation!5(3y+5)However if you are factorising quadratic expressions the method is similar but slightly similair. You need to find the factors of the last number that will make the sum of the middle number. I know your thinking it's difficult but once you know how to do it, it's easy!p2 - 8p + 15 = (p-3)(p-5)1. Find the factors of 1515 x 13 x 52. Find the two which will sum up to 83 x 53. Use the squared number/letter in the expression(p-3)(p-5)It can get harder be aware, so if you need more help I suggest youu search the internet.Answer BJust to correct that the above answer is wrong for factorising 15y + 10.The answer is 5(3y + 2). You have added 5 and 5 instead of timesing 5 and 2 which should give 10.Also the example that you gave for p2 - 8p + 15 = (p-3)(p-5) wasn't correct but the answer was.1. Firstly we find the factors of 15 such as:5x315x1-5x-3these all equal to 15.2. Then we find the factors of 15 that add up to -8which are -5 and -33. Now we put it all together. P² will set up as (p )(p ) this means p² in expanding brackets.Now it becomes: (p-5)(p-3) if we expand this answer which is doing the opposite of factorising, the answer will become back to p2 - 8p + 15.
What does factor mean in algebra?
When you factor, you take a common polynomial and multiply it by another polynomial (both with the same variable [i.e. both having "y"]) which when solved would create the polynomial in the problem.1) -p2+8p-12-(p-6)(p-2)2) 2y2+15y+7(y+7)(2y+1)3) 6c2+7c+2(2c+1)(3c+2)4) -2b2-5b+12-(b+4)(2b-3)
