No.
3
a=s/t, and s=d/t, so if we substitute... a = (d/t)/t --> a = d/t2 You must know both the acceleration and time in order to solve for the distance travelled.
You can solve for a one-time constant by using the formula t = RC. Read the math problem you are given carefully to determine what values to plug into the equation.
The phrase "triangle plus square equals 13" suggests a mathematical expression where a triangle and a square represent unknown values that, when added together, equal 13. If we assign variables to these shapes, such as ( T ) for triangle and ( S ) for square, the equation can be written as ( T + S = 13 ). To solve for one variable, we would need additional information or equations involving ( T ) and ( S ). Without that, there are multiple combinations of values for ( T ) and ( S ) that satisfy the equation.
You can't "solve" since this isn't "=" to anything.
3
S=vt-16t2 solve for v is what I will assume you mean. first pull out the t S=t(v-16t) then devide by t S/t=v-16t Then add 16t to both sides S/t + 16t = v This can also be written as (S+16t2)/t = v
you take a s**t
P=s r t , so, s= P/(st)
SOLVE : 21t + t=
4 solve t
f(t)dt and when f(t)=1=1/s or f(t)=k=k/s. finaly can be solve:Laplace transform t domain and s domain L.
S = 3 st + 3t = 6 st = 6 - 3t s = (6 - 3t)/t = 6/t - 3
a=s/t, and s=d/t, so if we substitute... a = (d/t)/t --> a = d/t2 You must know both the acceleration and time in order to solve for the distance travelled.
Given T = R + RS Lateral inversion makes it to be R + RS = T Taking R as common factor, we get R(1+S) = T Now dividing by (1+S) both sides, R = T / (1+S) Hence the solution R = T/(1+S)
it´s impossible, don´t still trying
If you have an expression with many variables, you can often solve for a different one. One classic example is rate x time = distance. We use the variable rxt=d Now this is solved for d, but say you want to solve for t. You divide both sides by r and t=d/r and it is solved for t.