Solving Equations Algebraically?

Question

1.) V=2*pie*r divided by t t= r=
(answer: t=2*pie*r divided by v),
(answer: v*t divided by 2 * pie)

2.) d= V sub0 * t + 1/2 * A * T square a= vsub0=
a= 2(d- V sub0 * t) diivided by t square
v sub0= (d-1/2*a*tsquare) divided by t

3.) t= 2 * pie * Square root of l/g l= g=
l= T square * g divided by 4 * pie square
g= 4 Pie square * l divided by T square

4.) a= 4 pie square * r divided by t square t= r=
t= 2 pie * square root of r/a
r= a * t square divided by 4 * pie square

5.) V sub f square = v sub 0 square + 2 a * d
v sub0 = vf square - square root of 2 a * d
d= v subf f square - 2 v sub 0 sauare divided by 2 a

6.) d= 1/2 (v sub 0 + v sub f) * t v sub f = t=
v sub f = 2 d - v sub 0 divided by t

Answer 1

1.)
V = (2pi * r)/t

Just switch the V and t and you get
t = (2pi * r)/V

Vt = 2pi * r
r = (Vt)/(2pi) or (1/2)((Vt)/pi)

t = (2pi * r)/V
r = (Vt)/(2pi)

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2.)
d = Vot + (1/2)at^2

d - Vot = (1/2)at^2
2d - 2Vot = at^2
a = (2d - 2Vot)/(t^2) or 2((d - Vot)/t^2)

d - (1/2)at^2 = Vot
Vo = (d - (1/2)at^2)/t

ANS :
a = 2((d - Vot)/(t^2))
Vo = (d - (1/2)at^2)/t

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3.)
t = 2pi * sqrt(l/g)

(t/(2pi) = sqrt(l/g)
(t/(2pi))^2 = l/g
l = (t/(2pi))^2 * g or (t^2 * g)/(4pi^2)

(t/(2pi))^2 = (l/g)
g = (l/((t/(2pi))^2)
g = (l/1)/((t^2)/(4pi^2)))
g = (l/1)*((4pi^2)/(t^2))
g = (4pi^2 * l)/(t^2)

ANS :
l = (t^2 * g)/(4pi^2)
g = (4pi^2 * l)/(t^2)

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4.)
a = (4pi^2 * r)/(t^2)

t^2 = (4pi^2 * r)/a
t = sqrt((4pi^2 * r)/a)
t = 2pi * sqrt(r/a)

at^2 = 4pi^2 * r
r = (at^2)/(4pi^2)

ANS :
t = 2pi * sqrt(r/a)
r = (at^2)/(4pi^2)

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5.)
Vf^2 = Vo^2 + 2ad

Vf^2 - 2ad = Vo^2
Vo = sqrt(Vf^2 - 2ad)

Vf^2 - Vo^2 = 2ad
d = (Vf^2 - Vo^2)/(2a)

ANS :
Vo = sqrt(Vf^2 - 2ad)
d = (Vf^2 - Vo^2)/(2a)

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d = (1/2)(Vo + Vf) * t

d/t = (1/2)(Vo + Vf)
2(d/t) = Vo + Vf
2(d/t) - Vo = Vf

d/((1/2)(Vo + Vf))
(d/1)/((1/2)(Vo + Vf))
(2d)/(Vo + Vf)

ANS :
Vf = (2d/t) - Vo
t = (2d)/(Vo + Vf)

Answer 2

1) v=[2(pi)r]/t solve for t?

easy, divide both sides by t, and then by v.

answer: t= [2(pi)r]/v

solve for r?

mutiply both sides by t: v*t=2(pi)r

divide both sides by 2pi: vt/2pi = r

2) let v, = v(sub0)

d=v,t+(at^2)/2 solve for a?

multiply both sides by 2: 2d=v,+at^2

subtract v,t from both sides: 2d-v,t=at^2

divide both sides by t^2: 2(d-v,t)/t^2 =a

solve for v,?

mutliply both sides by 2: 2d=v,t+at^2

subtract at^2 from both sides: 2d-at^2=v,t

divide both sides by t: (2d-at^2)/t = v,

(same as your answer, without the fractions.)


so, I've done two, step-by-step for you. These are physics equations for position, velocity, acceleration, etc. If you are being exposed to these equations in your class, but can't even do the algebra to solve them, you should probably go back a class or two. Physics is calulus based, and calculus is a more complicated version of algebra...be sure you are ready for the class you are in, otherwise it may be too difficult and you may become discouraged.

Answer 3

Some of these are Physics equations.

1. Find the equation where you know all the variables except one.
2. Use your algebraic knowledge to solve for that variable.
3. Use your newfound answer in the other equations until you get the variable you initially sought.

Otherwise, this is gibberish to me. What do you need solved?

Answer 4

Is that like cherry pie, or apple ? I like pecan the best.
If you meant 3.14 that's "pi"

Answer 5

what is your question?