The question ask to find and compute the possible formula for the function below where as A = (-1,32) and B= (1,8). In my own calculation and further computation about the problem and also following the rule of general exponential function, the formula would be y = a*(b^x). I hope this would help
Answer:
<em>A=3 and B=6</em>
Step-by-step explanation:
<u>Increasing and Decreasing Intervals of Functions</u>
Given f(x) as a real function and f'(x) its first derivative.
If f'(a)>0 the function is increasing in x=a
If f'(a)<0 the function is decreasing in x=a
If f'(a)=0 the function has a critical point in x=a
As we can see, the critical points may define open intervals where the function has different behaviors.
We have
Computing the first derivative:
We find the critical points equating f'(x) to zero
Simplifying by -6
We get the critical points
They define the following intervals
Thus A=3 and B=6
Answer:
2
Step-by-step explanation:
Answer:
27
Step-by-step explanation:
135 divided by 5 (divide both sides by 5)
9514 1404 393
Answer:
1a: x+3 = 5
1c: 6 = 2z
2b: x = 2
2d: 3 = z
3: the solutions make the hangars balance
Step-by-step explanation:
1. We can write the equations by listing the contents of the hangar and using an equal sign to show the balance between left side and right side. It can work well to put left side contents of the hangar on the left side of the equal sign.
A: x + 3 = 5
C: 1 + 1 + 1 + 1 + 1 + 1 = z + z simplifies to 6 = 2z
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2. B: We can subtract 3 from both sides of the hangar (and equation) to find the value of x.
(x +3) -3 = 5 -3
x = 2 . . . . . hangar balances with 2 on the right
D: We can divide both sides of the hangar by 2, splitting the content into two equal parts. Then one of those parts can be removed from each side.
2(3) = 2(z)
3 = z . . . . . . hangar balances with 3 on the left
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3. The found values will keep the hangar in balance when they are substituted for the corresponding variables.
A: 2 + 3 = 5
C: 1 + 1 + 1 + 1 + 1 + 1 = 3 + 3
