for each pair of expressions below, without substituting in specific values, determine which of the expressions in the given pairs is larger. explain your reasoning in a sentence or two. if you need to, test with values, but your explanation should explain what's going on without simply describing the values
you plugged in.

1) 5 + t² will always be larger than 3 - t².

2) 15/x² is larger than 15/7x²

3) 1/x will be larger than 1/(-x), only if the values of x take positive numbers.

4) k² will be larger than the second (k-6)². if k takes only positive numbers.

1)

If we see the first term, 5 is adding the function t² and in the second term,  t² is negative so we can conclude that 5 + t² will always be larger than 3 - t².

2)

Here, we can analyze the independent values. The first term has 15 and the second 15/7 = 2.14. Now, as we have x² in the bottom, the function decrease as x increase. Therefore, the first term 15/x² is larger than the second 15/7x²

3)

We can see that the second term is negative, so 1/x will be larger than 1/(-x), only if the values of x take positive numbers.

4)

Here we have the same case, if k takes only positive numbers, the first term k² will be larger than the second (k-6)² because we have 6 substracting.

You can learn more about quadratic function here:

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I hope it helps you!  

Step-by-step explanation:

1. t² is a positive number. Adding a positive number to 5 will always produce a larger result than subtracting the same positive number from 3. The larger expression is 5+t².

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2. The expressions are only defined for x ≠ 0, so for x² a positive number. For any x, the expressions are both positive. 15/(7x²) is 1/7 of 15/x², so will always be smaller. The larger expression is 15/x².

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3. As in 2, these expressions are only defined for x ≠ 0. One expression is the opposite of the other. A number is greater than its opposite when it is positive, so 1/x > 1/-x for x > 0; and 1/-x > 1/x for x < 0.

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4. The expression (k -6)² has the same range of values as k², but its graph is shifted 6 units to the right. The left branch of (k -6)² will be greater than k² for any k < 3. Similarly, the right branch of k² will be greater than (k -6)² for any k > 3.

step-by-step explanation:

the answer is 128