In arithmetic sequence, let the first tern of the arithmetic sequence be, a, and the common difference, d, then the nth term, Tn, of the arithmetic sequence is given by:
For a linear function with y-intercept, c, and slope, m, the linear function is given by:
Comparing the equation of the arithmetic sequence and that of the linear function, we can see that y is compared to Tn, a is compared to c, m is compared to d, and x is compared to n - 1.
Therefore, <span>the common difference in an arithmetic sequence is like the slope of a linear function as both are multiple of a variable.</span>
Answer:
<u>Linear relationship</u>: increasing or decreasing one variable will cause a corresponding increase or decrease in the other variable.
<u>Inverse relationship</u>: the value of one variable decreases as the value of the other variable increases.
<u>Exponential relationship</u>: a constant change in the independent variable (x) gives the same proportional change in the dependent variable (y)
<u>Question 5</u>
As the x-value increases (by one unit), the y-value decreases.
Therefore, this is an inverse relationship.
The y-values are calculated by dividing 35 by the x-value.
**I believe there is a typing error in the table and that the y-value of x = 3 should be 11.67**
<u>Question 6</u>
As the x-value increases, the y-value increases. The y-value increases by a factor of 5 for each x-value increase of 1 unit.
Therefore, this is an exponential relationship.
Answer:
Neither linear nor exponential
Step-by-step explanation:
To check for a linear relationship. Find slope.
slope= (-1 - (-2)) / ( 5 - 2) = 1/3
check other points
slope = (1 - (-1) )/ (8 - 5) = 2/3
check more
slope = (4 - 1) / (11 - 8) = 3/ 3 = 1
Nope.
try assuming an exponential:
y = c * (a^x)
-2 = c* (a^2); -2/c = a^2
-1 = c *(a ^5); -1/c = a^5
1 = c * (a^8), 1/c = a^8
(-2/c)^4 = a^8 = 1/c
16/(c^4) = 1/c
c^3 = 16, then a = root (-2/ cube-root(16) )
The change from negative to postive would not work for y = c(a^x)
so...
assume y = a^x + k
-2 = a^2 + k
-1 = a^5 + k
... I would say neither..
The parent function is f(x) = x^3 with domain = all real numbers and range = all real numbers.
The given function is f(x) = x^3 - 2 with domain = all real numbers and range = all real numbers.
The answer to this question is two hundred and eighty
