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Q: What is the solution to the equation 0 equals r?

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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.

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.

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

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.

A circle centre (0, 0) and radius r has equation x² + y² = r² The circle x² + y² = 36 has: r² = 36 → radius = 6

Let y=ce^(rx). R^2+r+1=0. Quadratic equation to find R.

13

It is an equation in the variable r.

25

It is an equation and the value of r is 66

p+c=r.

Solving equations in two unknowns (r and h) requires two independent equations. Since you have only one equation there is no solution.

r = 25

cp-cv =R proved that//

r = 4

r2 + r - 20 = 0(r + 5)(r - 4) = 0r + 5 = 0 or r - 4 = 0r = -5 or r = 4

Solution:(5r0)2 - (1r1+0r0)*5 + 3r1+1r0 = (8r0)2 - (1r1+0r0)*8 + 3r1+1r0 25 - 5r + 3r + 1 = 64 - 8r + 3r + 13r - 39 = 0r = 13Result: radix r = 13

113.04 - 113.1

300

If p, q, r, ... are the roots of the equations, then (x-p), (x-q), (x-r), etc are the factors (and conversely).

(2-r)e-rr

The question is rather vague but the answer to it could be:- p = 9 and r = 2

r + 14 = 39 r = 25