How do I rearrange equations?
Simple you add or subtract the value you want to move trivial example:[math]2+3=5[/math]You could rearrange it by writting:[math]2=5-3[/math]Or[math]2+3-5=0[/math]Or[math]3=5-2[/math]You can add something completely new if it helps you simplefy things:[math]2+3+c=5+c[/math]Also[math]a=b[/math]can be rearanged to[math]b=a[/math]
How can you rearrange SUVAT equations?
The SUVAT equations are:Now with these equations rearranging isn’t too difficult. You should just remember that whatever you do to one side of the equation, you have to do to the other.The only part you might slip up on is remembering that when you have to undo a square root (for example the one attached to our time, [math]t[/math]), you will have to have a [math]\pm[/math] sign at the start of your result since you end up with two answers when using a square root (since [math]x^2[/math] has two roots).I’ll do an example question, let’s say I had to isolate our acceleration, [math]a[/math], in the fourth equation:[math]v^2 = u^2 + 2as[/math][math]v^2 - u^2 = 2as[/math][math]\frac{v^2 - u^2}{2} = as[/math][math]\frac{v^2 - u^2}{2s} = a[/math]Or I could be tasked with isolating our time variable in the third equation:[math]s = vt - \frac{1}{2} at^2[/math][math]s - vt = - \frac{1}{2} at^2[/math][math]vt - s = \frac{1}{2} at^2[/math][math]2(vt - s) = at^2[/math][math]\frac{2(vt - s)}{a} = t^2[/math][math]\pm \frac{\sqrt{2(vt -s)}}{\sqrt{a}} = t[/math]All problems on rearranging are just variations of this, just be sure to do the same operations on both sides and not forgetting to do it on one.
What is the rate equation if rearranged?
The equation Rate=k(a)(b) if you were to rearrange it to solve for (b) or (a) what would it be? and somebody explain how to calculate % ionization? %I=x/acid base x100 ??
How to rearrange this linear equation?
Y = a + bX Subtract the a from both sides. Y - a = bX Next, divide the b from both sides. (Y - a) / b = X Which is the same as: X = (Y - a) / b
How do I rearrange the equation to make m the subject?
Question: P*t = m*c*deltaT + m*LP*t = m*(c*deltaT + L) ————————->Distributive propertyThis is the crux of finding m. PRACTICE THIS. It may be the single most useful basic algebraic property you will ever use, alongside multiplying by one and adding zero.(P*t)/(c*deltaT + L) = m ——————— > divide both sides by (c*deltaT + L)I assume this is the equation you’re looking for.m = (P*t)/(c*deltaT + L)
Rearranging equation help?
I would say that it makes perfect sense look Ka=[H+][A-]/[HA] You will multiply HA with Ka. therefore (Ka x [HA])/[A-] = [H+] The [A-] came below (Ka x [HA]) because algebrically we divide both sides by [A-] Got it?
How do I rearrange equations in chemistry?
it is just simple multiplying and divding. If you are trying to isolate a variable on one side either multiply or divide out all of the other terms to the opposing side. Take this for example A*B*C = D*E*F (solve for A) A = (D*E*F)/(B*C) Or Take (A*B)/(C*D)=(E*F)/(G*H) ----> Solve for D 1. (A*B)/C = D* (E*F)/(G*H) 2. (A*B*G*H)/(C*E*F) = D Just push the variables around
How do i rearrange this physics equation?
a = Dv / Dt In this equation a = acceleration, Dv = change in velocity, and Dt = change in time. If the acceleration of the object is the same for all time intervals, the acceleration then is said to be constant or uniform. Acceleration is measured in meters per second per second or m/s/s, or m/s2. Other equations that are related to acceleration are: Velocity of an object with constant acceleration: vf = vi + at In the above equation vf = final velocity of the object, vi = initial velocity of the object, a = acceleration, and t = time. Displacement during constant acceleration: d = 1/2(vf + vi)t The variables in the above equation are the same as in the previous equation. Displacement when acceleration and time are known: d = vit + 1/2at2 The above equation is composed of two terms. The first term, vit, corresponds to the displacement of an object if it were moving with constant velocity, vi. The second term, 1/2at2, gives the displacement of an object starting from rest and moving with uniform acceleration. The sum of these two terms gives displacement of an object that starts with an initial velocity and accelerates constantly. Displacement when velocity and acceleration are known: vf2 = vi2 + 2ad
Can someone try to rearrange this equation, q=t/(1-t) [ to make t the subject?
q = t/(1 - t) (Given)Now, multiply both sides of the equation by 1 - t to clear the equation of fractions involving t and to begin isolating t on one side as follows:q(1 - t) = [t/(1 - t)](1 - t)q(1 - t) = [t/(1 - t)][(1 - t)/1]Multiplying fractions on the right, we get:q(1 - t) = [t(1 - t)]/[(1 - t)(1)]q(1 - t) = (t/1)[(1 - t)/(1 - t)]q(1 - t) = (t/1)(1)q(1 - t) = tNow, using the Distributive Property on the left side as follows:q - qt = tNow, add qt to both sides as follows:q - qt + qt = t + qtq + (-qt) + qt = t + qtq + [-qt + qt] = t + qtq + 0 = t + qtq = t + qtNow, factoring out t on the right side as follows:q = (1 + q)tNow, dividing both sides by 1 + q to isolate and therefore solve for t:q/(1 + q) = [(1 + q)t]/(1 + q)q/(1 + q) = [(1 + q)/(1 + q)](t/1) q/(1 + q) = [1]tq/(1 + q) = t ort = q/(1 + q) since equality is symmetric, i.e., if a = b, then b = a.CHECK:q = t/(1 - t)q = [q/(1 + q)]/[1 - (q/(1 + q))] q = [q/(1 + q)]/[(1 + q)/(1 + q) - (q/(1 + q))] q = [q/(1 + q)]/{[(1 + q) - q]/(1 + q)}q = [q/(1 + q)]/[(1 + 0]/(1 + q)]q = [q/(1 + q)]/[1/(1 + q)]q = [q/(1 + q)][(1 + q)/1]The (1 + q) factors cancel out on the right, leaving us:q = (q/1)(1/1)q = qTherefore, t indeed equals q/(1 + q).
How can I rearrange the equation s=ut+1/2at^2 to make u the subject?
The equation [math]s=ut+\displaystyle\frac{at^2}{2}[/math] is a well-known equation of motion for constant acceleration. It is used to find the displacement ([math]s[/math]) by having the initial velocity ([math]u[/math]), the acceleration ([math]a[/math]) and the time taken ([math]t[/math]). However, if we have the acceleration and need to find the initial velocity, we could rearrange the equation. This involves isolating the [math]u[/math] and making it the subject.[math]s=ut+\displaystyle\frac{at^2}{2}[/math]Flip the equation around so that the term with a [math]u[/math] in it is on the left hand side (LHS) - this is my personal preference and is not necessary, but do it if you want[math]ut+\displaystyle\frac{at^2}{2}=s[/math]We need to get the [math]u[/math] by itself. We can subtract the fraction from both sides to get rid of one term.[math]ut=s-\displaystyle\frac{at^2}{2}[/math]All we need to do now to get the [math]u[/math] by itself is to get rid of the [math]t[/math]. This is easy, we can divide both sides by [math]t[/math][math]u=\displaystyle\frac{s}{t}-\displaystyle\frac{at^2}{2t}[/math]We can simplify the fraction by cancelling a [math]t[/math] and we get our final equation:[math]u=\displaystyle\frac{s}{t}-\displaystyle\frac{at}{2}[/math]