Step-by-step explanation:
x(x-3)=0
x=0 or (x-3) =0
x= 0 or x =3
it's plural so axes*
and in pilar coordinates, the axes are circular so they never intersect but the centre point is still called origin.
C. f(x) = – 2 cos 6x + 1
Start by determining the amplitude. Since we've deduced the amplitude is 2, the equation can include either a positive or negative 2 (since amplitude measures absolute value).
Next is the period. The equation for period P is P = (2pi)/b. If P is pi/3, then
pi/3 = (2pi)/b. Thus your b value should be 6.
Finally, the midline would be given by + 1 since adding a unit shifts the function upwards. This means that instead of the highest y value being 2 and the lowest -2, instead you'd have values of 3 and -1.
(3 – 1)/2 = 1 (midpoint theory).
All the numbers in this range can be written as

with

and

. Construct a table like so (see attached; apparently the environment for constructing tables isn't supported on this site...)
so that each entry in the table corresponds to the sum of the tens digit (row) and the ones digit (column). Now, you want to find the numbers whose digits add to perfect squares, which occurs when the sum of the digits is either of 1, 4, 9, or 16. You'll notice that this happens along some diagonals.
For each number that occupies an entire diagonal in the table, it's easy to see that that number

shows up

times in the table, so there is one instance of 1, four of 4, and nine of 9. Meanwhile, 16 shows up only twice due to the constraints of the table.
So there are 16 instances of two digit numbers between 10 and 92 whose digits add to perfect squares.
Answer:
Step-by-step explanation:
Given problem: C(x,y) = 36x + 48y
constraint: 100x^0.6y^0.4
Using langrange Multiplier,
36 = 0.6(100)x^-0.4y^0.4λ i
48 = 0.4(100)x^0.6y^-0.6λ ii
dividing the equations we have:
x = 2y
substituting into the constraint
p(x,y) = 100 *(2y)^0.6 y^0.4 = 100*2^0.6 *y
5000 = 151.572y
y = 329.876 labor units
x = 659.752 capital units
Minimum cost = 36(659.752) +48(329.876) = $39585.12