We know that
When we have

Here ,
a is the leading coefficient
c is the constant term
so, we can compare it with

so, we get

so,
leading coefficient is 5
constant term is 6......................Answer
Answer:
b = -125/9
Step-by-step explanation:
1 + log5 ( -9b) = 4
Subtract 1 from each side
1 -1+ log5 ( -9b) = 4-1
log5 ( -9b) = 3
Raise each side to the base 5
5^ log5 ( -9b) = 5^3
This will cancel the log5
-9b = 125
Divide each side by 9
-9b/-9 = 125/-9
b = -125/9
The answer is 2 buckets
Explain:
Tim picked 1 7/12 and Fred picked 4 2/3
These as improper fractions are 19/12
And 8/3
To make these have the same denominator you would times the 8/3 by 4 to give 32/12
Take these two away, Fred has 13 more apples.
And as we calculated it with there being 12 in each bucket there would be one full bucket but we would need another bucket for the apple left
So Fred picked 2 more buckets than Tim.
Hoped it makes sense and that I helped
Answer:
The domain and range remain the same.
Step-by-step explanation:
Hi there!
First, we must determine what increasing <em>a</em> by 2 really does to the exponential function.
In f(x)=ab^x, <em>a</em> represents the initial value (y-intercept) of the function while <em>b</em> represents the common ratio for each consecutive value of f(x).
Increasing <em>a</em> by 2 means moving the y-intercept of the function up by 2. If the original function contained the point (0,x), the new function would contain the point (0,x+2).
The domain remains the same; it is still the set of all real x-values. This is true for any exponential function, regardless of any transformations.
The range remains the same as well; for the original function, it would have been
. Because increasing <em>a</em> by 2 does not move the entire function up or down, the range remains the same.
I hope this helps!