8a+12b because you multiply 2 by each number. For example 4a times 2 is 8a and 6b times 2 is 12b
Hope that helped :)
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.
That would be 6 *3 - 8*0 = 18
Its D
Answer: x^2 + y^2 -10y = 0
Step-by-step explanation:
Cartesian coordinates, also called the Rectangular coordinates, isdefined in terms of x and y. So, for the problem θ has to be eliminated or converted using basic foundations that are described by the unit circle and the right triangle trigonometry.
r= 10sin(θ)
Remember that:
x= r × cos(θ)
y= r × sin(θ)
r^2= x^2 + y^2
Multiply both sides of the equation by r. This will give:
r × r = 10r × sin(θ)
r^2 = 10r × sin(θ)
x^2 + y^2= 10r × sin(θ)
Because y= r × sin(θ), we can make a substitution. This will be:
x^2 + y^2= 10y
x^2 + y^2 -10y = 0
The above equation is the Rectangular coordinate equivalent to the given equation.