-9r plus 10r can be written as ( -9r + 10r) and the answer is r.
(r + 5)(r + 5)
21
10r2 - 29r + 10 = (5r - 2)(2r - 5) when factored Work: First, the trinomial is factorable because ac = 100 = (-25)(-4) and b = -29 = (-25) + (-4). I would like to multiply by 10/10 the trinomial: 10r2 - 29r + 10 = (10/10)(10r2 - 29r + 10) = [(10r - 4)(10r - 25)]/(2)(5) = (5r - 2)(2r - 5)
10r(2)/r(8)= 9.17632039814324000 with the zeroes repeating. To figure out problems like this, you can go to Microsoft excel and enter formulas. That is what I did to figure out this problem. In this case, the formula would "=(10^sqrt(2))/sqrt(8).
-9r plus 10r can be written as ( -9r + 10r) and the answer is r.
The quadratic expression: r2+10r+25 = (r+5)(r+5) when factorised
-9r = 20 + rsubtract r from both sides-10r = 20divide both sides by -10r = -2
(r + 5)(r + 5)
so -10r = 2 so r = -0.2
Divide all terms by 5 and then it factors out as: (2r-9)(r+6)
i have a Johnson MQ-10R it is a 1964 model
the answer is -10r
-180
it cant be reduced
21
The Evinrude outboard, model MQ-10R, would be a 1964 model, 9.5 hp.