The Answer Will Be 1.1205 x 10 To The 5th Power.
Answer:
D. y=30x
Step-by-step explanation:
So x represents like say if he mowed 3 yards.
y=30(3)
Then y would equal 90
Answer: C
<u>Step-by-step explanation:</u>
The general form of the equation is: g(x) = a|x - h| + k ;
- "a" represents the vertical stretch <em>(or shrink)</em>
- "h" represents the x-coordinate of the vertex (left and right)
- "k" represents the y-coordinate of the vertex (up and down)
4 units left means h = 4
2 units up means k = 2
--> g(x) = |x - 4| + 2
Answer:
x = 20 in
L = 40 in
Step-by-step explanation:
Solution:-
- Denote the following:
The side of square cross section = x
The length of package = L
- Given that the combined length "L" of the package and girth "P" of the package must be less than and equal to 120 in
- The girth of the package denotes the Perimeter of cross section i.e square:
P = 4x
- The constraint for our problem in terms of combined length:
L + 4x ≤ 120
L = 120 - 4x .... Eq1
- The volume - "V" -of the rectangular package with a square cross section is given as:
V = L*x^2 ... Eq2
- Substitute Eq1 into Eq2 and form a single variable function of volume "V":
V(x) = 120*x^2 - 4x^3
- We are asked to maximize the Volume - " V(x) " - i.e we are to evaluate the critical value of "x" by setting the first derivative of the Volume function to zero:
d [ V(x) ] / dx = 240x - 12x^2
240x - 12x^2 = 0
x*(240 - 12x) = 0
x = 0, x = 20 in
- We will plug in each critical value of "x" back in function " V(x) ":
V (0) = 0
V(20) = 120(20)^2 - 4(20)^3
= 16,000 in^3
- The maximizing dimension of cross section is x = 20 in, the length of the parcel can be determined by the given constraint Eq1:
L = 120 - 4*20
L = 40 in
- The maximum volume of the rectangular package is with Length L = 40 in and cross section of Ax = ( 20 x 20 ):
Answer:
is there a graph to go with it ?
Step-by-step explanation:
