How do you solve for t in # s/t= z/v#?

Answer 1

#t=(vs)/z#

#color(blue)("I am showing you the 'first principle' method upon which the")##color(blue)("'shortcut' method is based.")#
It is perfectly allowed to turn the whole thing upside down. That way you get the #t# as a numerator (the top number of a fraction).
#t/s=v/z#
Now we need to get #t# on its own on one side of the equals and everything else on the other side.
So we need to 'get rid' of the #s#. For multiply or divide we do this by changing it to 1. If it was add or subtract we would change it to 0.
Multiply both sides by #s#
#s xx t/s=v/z xx s#

This is the same as

#s/s xx t=(vs)/z#
But #s/s =1#
#1 xx t =(vs)/z#
But #1xx t = t#
#t=(vs)/z#
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Answer 2

To solve for ( t ) in ( \frac{s}{t} = \frac{z}{v} ), you can cross multiply to isolate ( t ):

[ s \times v = t \times z ]

Then, divide both sides of the equation by ( z ) to solve for ( t ):

[ t = \frac{s \times v}{z} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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