The school choir earned $5,012 selling tickets to their winter concerts. They sold 324 adult tickets for $10. 75 each. If studen
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Answer:
you're not doing anything wrong
Step-by-step explanation:
In order for cos⁻¹ to be a function, its range must be restricted to [0, π]. The cosine value that is its argument is cos(-4π/3) = -1/2. You have properly identified cos⁻¹(-1/2) to be 2π/3.
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Cos and cos⁻¹ are conceptually inverse functions. Hence, conceptually, cos⁻¹(cos(x)) = x, regardless of the value of x. The expected answer here may be -4π/3.
As we discussed above, that would be incorrect. Cos⁻¹ cannot produce output values in the range [-π, -2π] unless it is specifically defined to do so. That would be an unusual definition of cos⁻¹. Nothing in the problem statement suggests anything other than the usual definition of cos⁻¹ applies.
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This is a good one to discuss with your teacher.
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