True
Hope this helps
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The given table is representing Profits for the given years. The independent variable is represented on x axis and on y axis dependent variable is taken .Years of increment is the independent variable and profit is the dependent variable .Year is represented on the x axis while profit will be represented on y axis. The appropriate profit in increments is $50,000
Among all the given options Option a)x-axis: years in increments of 1; y-axis: profit in increments of $50,000 is the correct answer.
Answer:
5
Step-by-step explanation:
-10x-1=-51
Add 1 to both sides:
-10x=-50
Divide both sides by -10:
x=5
Hope this helps!
Answer:
The maximum profit is 5070 dollars
Step-by-step explanation:
The profit y is represented by a quadratic function.
The equation that we can use to find a maximum value of a quadratic function is:
Max_value = c - (b^2 / 4a)
Where a, b and c are the coefficients of the quadratic function (in our case: a = -13, b = 770, c = -6332)
So, using this equation, we have:
Max_value = -6332 - (770^2 / 4*(-13)) = -6332 - 592900/(-52)
Max_value = -6332 + 11401.92 = 5069.92
rounding to the nearest dollar, we have that the maximum profit is 5070 dollars.
