86 + 86 x 5 = 516
18 + 18 x 5 = 108
516 + 108 = 624
Answer:
C(x) = 5700 + 500x
Step-by-step explanation:
Total cost = fixed cost + variable cost
Fixed cost are cost that do not change in the process of production such as machineries, furnitures & fittings.
Variable cost are cost that changes with production such as cost of raw material, cost of labor.
From the question:
Fixed cost= $5,700
Cost per unit of computer = $500
If x number of computers are produced in a month
Variable cost = 500x
Total cost = fixed cost + variable cost
C(x) = 5700 + 500x
Where,
C(x) = total cost
5700= fixed cost
500x= variable cost
Given: x^2 - 6x + 2
Statements:
1) The graph of the quadratic equation has a minimum value:TRUE. WHEN THE COEFFICIENT OF X^2 IS POSITIVE THE PARABOLA OPEN UPWARDS AND ITS VERTEX IS THE MINIMUM.
2) The extreme value is at the point (3 , - 7): TRUE
You have to find the vertex of the parabola:
x^2- 6x + 2 = (x - 3)^2 - 9 + 2 = (x - 3)^2 - 7 => vertex = (3, -7)
3) The extreme value is at the point (7, -3): FALSE. THE RIGHT VALUE WAS FOUND IN THE PREVIOUS POINT.
4) The solutions are x = - 3 +/- √7. FALSE.
Solve the equation:
(x - 3)^2 - 7 = 0 => (x - 3)^2 = 7 => (x - 3) = +/- √7 => x = 3 +/- √7
5) The solutions are x = 3 +/- √7. TRUE (SEE THE SOLUTION ABOVE).
6) The graph of the quadratic equation has a maximum value: FALSE (SEE THE FIRST STATEMENT).
Rewriting an expression so that it has no grouping symbols and all of the like terms have been combined is called "<span>Simplify An Expression".
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You have a larger right triangle (vertices: football field, park, home) with an altitude drawn to the hypotenuse. The altitude creates two more triangles. The two new triangles are also right triangles, and all three triangles are similar.
Part a.
Let x = distance from park to library
9/x = x/12
x^2 = 9 * 12
x^2 = 108
x = sqrt(108)
x = sqrt(36 * 3)
x = 6sqrt(3)
Answer to part a. 6sqrt(3)
Part b.
Once you know x, use the value of x and 12 as the lengths of legs, and you are looking for the distance from the park to the football field which is the hypotenuse.
Let y = the length of the hypotenuse which is the distance from the park to the football field.
(6sqrt(3))^2 + 12^2 = y^2
108 + 144 = y^2
y^2 = 252
y = sqrt(252)
y = sqrt(4 * 9 * 7)
y = 6sqrt(7)
Answer to part b: 6sqrt(7)
