We are given the equations:
<span>5 x + 11 y = 49 -->
eqtn 1</span>
<span>u = 3 x^2 + 6 y^2 -->
eqtn 2</span>
Rewrite eqtn 1 explicit to y:
11 y = 49 – 5 x
<span>y = (49 – 5x) / 11 -->
eqtn 3</span>
Substitute eqtn 3 to eqtn 2:
u = 3 x^2 + 6 [(49 – 5x) / 11]^2
u = 3 x^2 + 6 [(2401 – 490 x + 25 x^2) / 121]
u = 3 x^2 + 14406/121 – 2940x/121 + 150x^2/121
u = 4.24 x^2 – 24.3 x + 119.06
Derive then set du/dx = 0 to get the maxima:
du/dx = 8.48 x – 24.3 = 0
solving for x:
8.48 x = 24.3
x = 2.87
so y is:
y = (49 – 5x) / 11 = (49 – 5*2.87) / 11
y = 3.15
Answer:
<span>George will choose some of each commodity but more y than
x.</span>
The English alphabet contains 26 letters (a, b, c, ...y, z).
The digits from 0 to 9 are a total of 10.
A keycode contains 2 letters, and 3 numbers, for example:
AB 598; MM 139; NT 498; ...
So there are 26 possible choices for the first letter, which can combined with any of the 26 possible choices for the second letter, so there are a total of
26*26=676 possible pairs of letters.
Similarly, the 10 possible choices for the first number, which can be combined with the 10 possible choices for the second number, and the 10 possible choices for the third number make a total of :
10*10*10=1,000 possible triples of numbers.
Any of the 676 possible pairs of letters can be combined with any of the possible 1,000 triples of numbers. This makes a total of
676*1,000=676,000 keycodes.
Answer: 676,000
Answer:
3
Step-by-step explanation:
Answer:
x = 7
Step-by-step explanation:
A, B, and C are collinear B is between A and C.
AC = AB + BC
2x + 1 = x + 4 + 2x - 10
2x + 1 = 3x - 6
2x - 3x = - 6 - 1
- x = - 7
x = 7
The square root of 63 is about 7.9, which is between 7 and 8
