This is an excellent practice for the solution of quadratic equations.
1*36=36 => (1,36)
2*18=36 => (2,18)
3*12=36 => (3,12)
4*9=36 => (4,9)
6*6=36 => (6,6)
9*4=36 => (9,4)
12*3=36 => (12,3)
18*2=36 => (18,2)
36*1=36 => (36,1)
We can see that the sum decreases until the two factors are close (or equal) and then increases again.
The pair of integers with a sum of 20 is therefore (2,18) or (18,2).
(2,3) so x=2, y=3 and h=(2^2+3^2)^(1/2)=√13
sina=3/√13, cosa=2/√13, tana=3/2
Afterwards multiply sina and cosa by √13/√13 and get sina=(3√13)/13 and cosa=(2√13)/13
Answer:-1/3
Step-by-step explanation: I really don’t know
Answer:
THE ANSWER IS -21
Step-by-step explanation: BECAUSE IF YOU TAKE -4P -15P=-19P
-19P-2R=-21
Answer:
Price of each phone: 
Price of each accessory: 
Step-by-step explanation:
Let be "p" the price in dollars of each phone and "a" the price in dollars of each accessory.
Set up a system of equations:

You can use the Elimination Method to solve the system.
Multiply the second equation by -2 and add both equations:

Substitute the value of "a" into the first equation and solve for "p":
