r=0 is the solution...
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∙ 2010-09-10 20:30:47It's a linear equation in 'r' . Here's how to find the solution:16 - r = -16Add 16 to each side of the equation:32 - r = 0Add 'r' to each side:32 = r
A circle centre (0, 0) and radius r has equation x² + y² = r² The circle x² + y² = 36 has: r² = 36 → radius = 6
It is an equation in the variable r.
25
p+c=r.
It is a linear equation in one variable r. The solution to the equation is r = 1.
5r = 25 is a linear equation in one variable, r. The solution is r = 5.
Divide both sides by 6: 3 - r = 0, so r = 3 satisfies the equation.
No, it is not. Substitute and check. You have 128 + 16r. So now, if r=9, then the equation would be 128 = 16x9. But, 16x9 is 145 and is not equal to 128. Thus, r=9 is not the solution. The solution is r+ 128/16 which is equal to 8. Thus, the actual solution is r = 8.
It's a linear equation in 'r' . Here's how to find the solution:16 - r = -16Add 16 to each side of the equation:32 - r = 0Add 'r' to each side:32 = r
A circle centre (0, 0) and radius r has equation x² + y² = r² The circle x² + y² = 36 has: r² = 36 → radius = 6
13
r + s = 4 and 2r + 3s = 8 multiply the first equation by 2 giving 2r + 2s = 8 subtract this from the second equation giving s = 0 So r = 4 and s = 0.
25
It is an equation in the variable r.
Let y=ce^(rx). R^2+r+1=0. Quadratic equation to find R.
100 r = 99.52Divide each side of the equation by 100 :r = 99.52/100 = 0.9952