Answer:
Function A has a greater initial value because the initial value for Function A is 6 and the initial vale for Function B is 3
Step-by-step explanation:
The data for Function A is presented here as follows;
The slope of the function, 'm', is given using any two points as follows;
m = (9 - 3)/((-1) - 1) = -3
The slope of the function = -3
The equation of function in point and slope form is given as follows;
y - 3 = -3·(x - 1)
The equation of Function A, is therefore, given as follows;
y = 3 - 3·x + 3 = 6 - 3·x
∴ y = 6 - 3·x
The equation of Function B is given as follows;
y = 6·x + 3
The initial value of a function is given by the y-intercept of the function, where the input variable (x-variable) is zero
The initial value, (y-intercept) of Function A, f(0) is therefore found as follows;
f(x) = y = 6 - 3·x
f(0) = 6 - 3×0 = 6
The initial value of Function A, f(0) = 6
Similarly, the initial value, (y-intercept) of Function B, f(0) is found as follows;
f(x) = y = 6·x + 3
f(0) = 6×0 + 3 = 3
The initial value of the Function B, f(0) = 3
Therefore, we have that Function A has a greater initial value than Function B
Answer with Step-by-step explanation:
Since we have given that
a + b = c
and a|c
i.e. a divides c.
We need to prove that a|b.
⇒ a = mb for some integer m
Since a|c,
So, mathematically, it is expressed as
c= ka
Now, we put the above value in a + b = c.
So, it becomes,
a=mb, here, m = k-1
Hence, proved.