When you divide 52 by 8, you have to find the nearest multiple of 8 that is either equal to or less than 52.
closest multiple of 8 that is less than 52 is 48. Quotient is the number of times 8 is in 48. In other words 8 is there 6 times in 48.
quotient is found by dividing the closest multiple of 8 (48) by 8.
48 / 8 = 6
quotient is 6
remainder is the difference between the number given and closest multiple
number given is 52. closest multiple is 48. difference between 52 and 48 is
52 - 48 = 4
the remainder is 4
quotient is 6
Answer:
x + 7 = y // y x 8
Step-by-step explanation:
x = ?
y + ?
x and y are the variables, since we don't know what the values are
Hope this helps!!
The answer is 2 I think.......
Answer:
(x, y) → (-x, -y)
Q"(4, 1) → Q'(-4, -1)
R"(6, -5) → R'(-6, 5)
S"(3, -4) → S'(-3, 4)
T"(1, -1) → T'(-1, 1)
Step-by-step explanation:
this might be right, sorry if im wrong
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 ):
