How do you evaluate #a-2y# when #a=-2/3# and #y=4/5#?

Answer 1

You will get the answer #-34/15#

Evaluating something is basically substituting a value in for a variable. They give you value -2/3 for a, and they give you 4/5 for y. So when we sub those in we get the equation

#-2/3 - 2(4/5)#

So this looks pretty easy to solve. From here, if we use PEMDAS we see that we must multiply to by the fraction first and we get

#-2/3-8/5#

Since we can't subtract without like denominators we want to find the lowest common denominator, which in this case will be 15. Now we have our new fractions

#-10/15-24/15#

So if we subtract the two numerators we get -34

#-34/15#

And this fraction can't be simplified so it is our final answer. Hope this helps!

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Answer 2

To evaluate ( a - 2y ) when ( a = -\frac{2}{3} ) and ( y = \frac{4}{5} ), substitute the given values into the expression:

[ a - 2y = \left( -\frac{2}{3} \right) - 2 \left( \frac{4}{5} \right) ]

[ = -\frac{2}{3} - \frac{8}{5} ]

To combine the fractions, find a common denominator, which is 15:

[ = -\frac{10}{15} - \frac{24}{15} ]

[ = -\frac{34}{15} ]

So, ( a - 2y ) when ( a = -\frac{2}{3} ) and ( y = \frac{4}{5} ) is ( -\frac{34}{15} ).

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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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