Step-by-step explanation:
tokun Janwar Andhera development Savina School drawing
We apply the concept of calculus here. You can determine the minimum or maximum in here if you take the derivative of a function and equate it to zero. The solution is as follows:
n = 20t² - 20t +120
dn/dt = 0 = 40t - 20 + 0
40t - 20 = 0
40t = 20
t = 20/40
t = 1/2 or 0.5
Therefore, the number of bacteria would be minimal at 0.5°C.
There no other figure but can take a property that is linear pair or linear pair axiom
Answer: 180.5
9.5
X19.0
=180.5
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)
