Answer:
To match each functions with the corresponding function formula when h(x) = 5 - 3x and g(x) = -3 x + 5.
1. k(x) = (3h - 5g)(x)
= = 3(5 - 3x)-5(-3 x + 5)
= = 15 - 9x + 15x - 25
K(x) = -10 + 6x
2. k(x) = (h - g)(x)
= = (5 - 3x) - (-3 x + 5)
= = 0
k(x) = 0
3. k(x) = (5g + 3h)(x)
= = 5(-3 x + 5)+3(5 - 3x)
= = -15x + 25 + 15 - 9x
k(x) = - 24x +40
4. k(x) = (3g + 5h)(x)
= = 3 (-3 x + 5) + 5(5 - 3x)
= = -9x + 15 +25 -15x
k(x) = -24x + 40
5. k(x) = (g + h)(x)
= = (-3 x + 5) + (5 - 3x)
k(x) = -6x + 10
6. k(x) = (5h - 3g)(x)
= = 5(5 - 3x) - 3(-3 x + 5)
= =25 - 15x + 9x - 15
k(x) = 10 - 6x
Since the grade of the numerator and the denominator is the same, then the limit exists and is distinct from 0. The limit of the expression is 4/7.
<h3>How to determine the limit of a rational expression when x tends to infinite</h3>
In this problem we must apply some algebraic handling and some known limits to determine whether the limit exists or not. The limit exists if and only if the result exists.




4/7
Since the grade of the numerator and the denominator is the same, then the limit exists and is distinct from 0. The limit of the expression is 4/7.
To learn more on limits: brainly.com/question/12207558
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Answer:
(q)




(r)





(s)





Step-by-step explanation:
Answer:
It's letter b
Step-by-step explanation:
I hope this help
Its the third and the fourth one