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lakkis [162]
3 years ago
14

Plss help me with my question​

Mathematics
1 answer:
Tamiku [17]3 years ago
6 0

Answer:

1. 45.76, 52.17, 54.2, 55.16, 56.78

2. 9.18, 9.2, 9.98, 10.54, 11.23

3. 0.01, 0.05, 0.067, 0.07, 0.09

4. 33.14, 33.15, 33.45, 33.52, 33.55

5. 3.45, 4.45, 4.9, 5.49, 6.5

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solmaris [256]

Answers:

  1. Exponential and increasing
  2. Exponential and decreasing
  3. Linear and decreasing
  4. Linear and increasing
  5. Exponential and increasing

=========================================================

Explanation:

Problems 1, 2, and 5 are exponential functions of the form y = a(b)^x where b is the base of the exponent and 'a' is the starting term (when x=0).

If 0 < b < 1, then the exponential function decreases or decays. Perhaps a classic example would be to study how a certain element decays into something else. The exponential curve goes downhill when moving to the right.

If b > 1, then we have exponential growth or increase. Population models could be one example; though keep in mind that there is a carrying capacity at some point. The exponential curve goes uphill when moving to the right.

In problems 1 and 5, we have b = 2 and b = 1.1 respectively. We can see b > 1 leads to exponential growth. I recommend making either a graph or table of values to see what's going on.

Meanwhile, problem 2 has b = 0.8 to represent exponential decay of 20%. It loses 20% of its value each time x increases by 1.

---------------------

Problems 3 and 4 are linear functions of the form y = mx+b

m = slope

b = y intercept

This b value is not to be confused with the previously mentioned b value used with exponential functions. They're two different things. Unfortunately letters tend to get reused.

If m is positive, then the linear function is said to be increasing. The line goes uphill when moving to the right.

On the other hand if m is negative, then we go downhill while moving to the right. This line is decreasing.

Problem 3 has a negative slope, so it is decreasing. Problem 4 has a positive slope which is increasing.

7 0
1 year ago
Can anyone help me with this? please.
victus00 [196]
Acute right complementary adjacent
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