How do you write the mixed expression #(h-3)/(h+5)-(h+2)# as a rational expression?

Answer 1

#-(h^2+6h+13)/(h+5)#

#(h-3)/(h+5)-(h+2)# #=(h-3)/(h+5)-(h+2)*(h+5)/(h+5)# #=[(h-3)-(h^2+7h+10)]/(h+5)# #=-(h^2+6h+13)/(h+5)#
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Answer 2

To write the mixed expression (h-3)/(h+5)-(h+2) as a rational expression, we need to find a common denominator for the two fractions. The common denominator is (h+5).

Multiplying the first fraction by (h+5)/(h+5), we get (h-3)(h+5)/(h+5).

Multiplying the second fraction by (h+5)/(h+5), we get (h+2)(h+5)/(h+5).

Combining the two fractions, we have ((h-3)(h+5) - (h+2)(h+5))/(h+5).

Expanding and simplifying the numerator, we get (h^2 + 2h - 15 - h^2 - 7h - 10)/(h+5).

Combining like terms in the numerator, we have (-5h - 25)/(h+5).

Therefore, the rational expression for (h-3)/(h+5)-(h+2) is (-5h - 25)/(h+5).

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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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