If y'' = cIntegrating we get: y' = cx + aSubs: x = 0, y' = 1 so a = 1Now y' = cx + 1Integrating we get: y = cx^2/2 + x + dSubs: x = 0, y = 5 so d = 5Now y = cx^2/2 + x + 5
F=9/5C+32..........Solve for C?
The first person above said that you must subtract 32 first, but that is not true. It's better to subtract the 32 first, because that might make your calculation easier, but you could easily multiply by 5/9 first, that is divide everything by 9/5 and you will still obtain the right answer. I'll perform the calculation both ways: F = (9/5)C + 32 F - 32 = (9/5)C C = (5/9)[F - 32] 2nd way: dividing by 9/5 first: F = (9/5)C + 32 (5/9)F = C + (32)(5/9) C = (5/9)F - (5/9)(32) = (5/9)[F - 32]
How do you solve for c? c-3=-2(x+6)?
You have one equation and two variables. That means you can *solve for c in terms of x* (but not for a numerical answer.) To do this, you must *isolate the variable c*. That means, get c on one side of the equation all by itself. You can perform any operation on the equation, as long as you do it on both sides. For example: c-3 = -2(x+6) I can add 3 to both sides, and the equation stays the same. c-3 + 3 = -2(x+6) + 3 I chose to add three because it cancels out the minus 3. So now I have c = -2(x+6) + 3 If you want, you can simplify the right side of the equation. Distributing the -2 gives c = -2x -12 + 3 for a final answer of c = -2x - 9 (If you were to assign a value to x, then you could solve for c. c = -2x - 9 is the equation of a line)
How do you solve 1/2ax=c?
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C=2(pi)r Solve for r?
This Site Might Help You. RE: C=2(pi)r Solve for r? How do you do this? C=2(pi)r. How would you solve for r? Thanks!
That is a function, not an equation. So it cannot be “solved”. To convert it to an equation you have to specify a value for Y and then calculate the corresponding value of x. The simplest case is if Y = 0, because you obtain the traditional quadratic equation with the only exception that x cannot be 0. However, if Y is not zero, the situation gets more complicated because you have the quadratic equation LESS Y*sqrt(x) in a range of real that has to exclude 0.For example, the solution of (Y=1, a=1, b=4, c=-2)x^2+4x-2-sqrt(x)=0isx is about 0,60320
How do you solve: Solve in C: x^4 - 1 = 0?
x^4 - 20x^2 + 19 = 0 substitute y = x^2: y^2 - 20y + 19 = 0 ingredient and resolve for y: (y-one million)(y-19) = 0 y = one million or y = 19 replace x^2 = y: x^2 =one million or x^2 = 19 for that reason, the ideas are: x = +OR- one million and x = +OR- sqrt(19)
How do you solve (A dot B)C?
Use: A = (a1)i + (a2)j + (a3)k B = (b1)i (b2)j + (b3)k C = (c1)i (c2)j + (c3)k With with subscripts for the 1's, 2's and 3's. And i, j and k being the x, y and z vectors. I'm assuming you know how to find the dot products and cross products of those. It's a bit tedius, but substitute those into (A•B)C and (A×C)×B and try to get them looking the same.
Remove the parentheses. Now start the problem.c + 4 - 3c - 2 = 0c is of the same variable so you can subtract 3c from c. Same case with 4 and 2, causing you to subtract 2 from 4. You will get -2c + 2.Since you have ended up with -2c+2=0, you subtract 2 from both sides, resulting in -2c=-2.Solve for c but dividing -2 on both sides. You will get c =1.I think it is important that you are well-versed in the laws of algebra in order for you to work with problems like these in the future. Here is a link below for you to look at in order to get started:Basic Rules of Algebra