Answer:
y= 48 + 0.1x
x= miles driven
Step-by-step explanation:
<u>We need to calculate the fixed and variable cost (per mile) of renting a car. To do that, we will use the high-low method:</u>
Variable cost per unit= (Highest activity cost - Lowest activity cost)/ (Highest activity units - Lowest activity units)
Variable cost per unit= (65 - 58) / (170 - 100)
Variable cost per unit= $0.1 per mile
Fixed costs= Highest activity cost - (Variable cost per unit * HAU)
Fixed costs= 65 - (0.1*170)
Fixed costs= $48
Fixed costs= LAC - (Variable cost per unit* LAU)
Fixed costs= 58 - (0.1*100)
Fixed costs= $48
y= 48 + 0.1x
x= miles driven
Answer:
x + 4y ≤ 15; y ≥ 0
Step-by-step explanation:
The graph doesn't do a very good job of modeling any of the given equations. However, the equations listed above seem the best fit.
The slope of the top (left) line is negative, so the equation will be of the form ...
x + 4y = something
When y=0, x=15, so the "something" is expected to be 15.
However, the line appears to go through points (6, 2) and (-2, 4). Both of these points are on the line x + 4y = 14.
The graph is shaded <em>below the line</em> so the values of x and y that are in the shaded area will add to <em>less than</em> 15 (or 14). Hence, the inequality will be ...
x + 4y ≤ something . . . . . part of the 3rd answer choice
The fact that the shading does not go below y=0 means the other limit is ...
y ≥ 0 . . . . . part of every answer choice.
If x is a real number such that x3 + 4x = 0 then x is 0”.Let q: x is a real number such that x3 + 4x = 0 r: x is 0.i To show that statement p is true we assume that q is true and then show that r is true.Therefore let statement q be true.∴ x2 + 4x = 0 x x2 + 4 = 0⇒ x = 0 or x2+ 4 = 0However since x is real it is 0.Thus statement r is true.Therefore the given statement is true.ii To show statement p to be true by contradiction we assume that p is not true.Let x be a real number such that x3 + 4x = 0 and let x is not 0.Therefore x3 + 4x = 0 x x2+ 4 = 0 x = 0 or x2 + 4 = 0 x = 0 orx2 = – 4However x is real. Therefore x = 0 which is a contradiction since we have assumed that x is not 0.Thus the given statement p is true.iii To prove statement p to be true by contrapositive method we assume that r is false and prove that q must be false.Here r is false implies that it is required to consider the negation of statement r.This obtains the following statement.∼r: x is not 0.It can be seen that x2 + 4 will always be positive.x ≠ 0 implies that the product of any positive real number with x is not zero.Let us consider the product of x with x2 + 4.∴ x x2 + 4 ≠ 0⇒ x3 + 4x ≠ 0This shows that statement q is not true.Thus it has been proved that∼r ⇒∼qTherefore the given statement p is true.
Answer:
4
Step-by-step explanation:
