A cell phone company charges $60.00 a a month for up to 1 gigabyte of data. The cost of additional data is $0.05 per megabyte. I
1 answer:
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we know that
<u>Standard Form</u> is also known as Scientific Notation, is a method of writing numbers that accommodates value excessively large or small to be suitably written in Standard Decimal Notation
in this problem we have
Remember that
so
<u>convert to standard form </u>
therefore
<u>the answer is</u>
Answer:
Conclusion:
The rate of change of function 1 = 3
The rate of change of function 2 = 5/3
- Hence, function 1 has a greater rate of change
The initial Value of function 1 = y = 2
The initial Value of function 2 = y = 3
- Hence, function 2 has a greater initial value.
Step-by-step explanation:
Function 1)
Determining rate of change for function 1:
x 1 2 3 4
y 5 8 11 14
Finding the rate of change or slope using the formula
Rate of change = m = [y₂-y₁] / [x₂-x₁]
Taking any two points, let say (1, 5) and (2, 8)
Rate of change = m = [8-5] / [2-1]
= 3/1
= 3
Therefor, the rate of change of function 1 = m = 3
using point-slope form to determine the function equation
y-y₁ = m (x-x₁)
where m is the rate of change or slope
substititng m = 3 and the point (1, 5)
y - 5 = 3(x - 1)
y - 5 = 3x-3
y = 3x-3+5
y = 3x + 2
Thus, equation of function 1 will be:
y = 3x + 2
Determining Initial Value for Function 1:
substituting x = 0 in the equation to determine the initial value
y = 3(0)+2
y = 0+2
y = 2
Therefore, the initial Value of function 1 will be: y = 2
Function 2)
Determining the rate of change for function 2:
Given the function 2
comparing with the slope-intercept form of a linear function
y = mx+b where m is the rate of change
so the rate of change of function 2 = m = 5/3
Determining Initial Value for Function 2:
substituting x = 0 in the equation to determine the initial value
Therefore, the initial Value of function 2 will be: y = 3
Conclusion:
The rate of change of function 1 = 3
The rate of change of function 2 = 5/3
- Hence, function 1 has a greater rate of change
The initial Value of function 1 = y = 2
The initial Value of function 2 = y = 3
- Hence, function 2 has a greater initial value.