Answer:
u = x tan(A) - sec(A) sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2) or u = sec(A) sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2) + x tan(A)
Step-by-step explanation:
Solve for u:
(x sin(A) - u cos(A))^2 + (x cos(A) + y sin(A))^2 = x^2 + y^2
Subtract (x cos(A) + y sin(A))^2 from both sides:
(x sin(A) - u cos(A))^2 = x^2 + y^2 - (x cos(A) + y sin(A))^2
Take the square root of both sides:
x sin(A) - u cos(A) = sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2) or x sin(A) - u cos(A) = -sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2)
Subtract x sin(A) from both sides:
-u cos(A) = sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2) - x sin(A) or x sin(A) - u cos(A) = -sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2)
Divide both sides by -cos(A):
u = x tan(A) - sec(A) sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2) or x sin(A) - u cos(A) = -sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2)
Subtract x sin(A) from both sides:
u = x tan(A) - sec(A) sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2) or -u cos(A) = -x sin(A) - sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2)
Divide both sides by -cos(A):
Answer: u = x tan(A) - sec(A) sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2) or u = sec(A) sqrt(x^2 + y^2 - (x cos(A) + y sin(A))^2) + x tan(A)
Si bi almota con te illama
Answer:
The like terms are x + 2x which is 3x
and the rest you cant combine
Step-by-step explanation:
For a two-digit number to be palindromic, the two digits must be the same.
Number of two-digit numbers with the two digits equal is 9. (i.e. 11, 22, 33, . . . )
Total number of two-digit numbers is 90.
Therefore, the <span>probability that a randomly chosen two-digit number is palindromic is 1 / 10 or 0.1.</span>
Answer: yes it does
2x= 8
x= 8/2
x=4.
