The Lagrangian is
with critical points where the partial derivatives vanish.
Substitute 
into the last equation and solve for 
:
Then we get two critical points,
We get an absolute maximum of 
at the second point, and an absolute minimum of 
at the first point.
Answer:
Step-by-step explanation:
Given that:
The differential equation; 
The above equation can be better expressed as:
The pattern of the normalized differential equation can be represented as:
y'' + p(x)y' + q(x) y = 0
This implies that:
Also;
From p(x) and q(x); we will realize that the zeroes of (x+2)(x-2)² = ±2
When x = - 2
Hence, one (1) of them is non-analytical at x = 2.
Thus, x = 2 is an irregular singular point.
No, because there are TWO values of y for a certain value of x
say x=5, then y=2 and y=-2
That does not satisfy the concept of the function.
Answer:
The exact value of tan(M) is 5/12 ⇒ answer (C)
Step-by-step explanation:
* Lets revise the trigonometry functions
- In ΔABC
# m∠B = 90°
# Length of AB = a , length of BC = b and length of AC = c
# The trigonometry functions of angle C are
- sin(C) = a/c ⇒ opposite side to ∠C ÷ the hypotenuse
- cos(C) = b/c ⇒ adjacent side to ∠C ÷ the hypotenuse
- tan(c) = a/b ⇒ opposite side to ∠C ÷ adjacent side to ∠C
* Now lets solve the problem
- In ΔONM
∵ m∠N = 90°
∵ MN = 12
∵ ON = 5
∵ tan(M) = ON/NM ⇒ opposite side of ∠(M) ÷ adjacent side of ∠(M)
∴ tan(M) = 5/12
* The exact value of tan(M) is 5/12
Answer:
3/3 is equivilant to 1.
Multiplying 6/4 by 1, would be 6/4. Also, multiplying 6/4 by 3/3 would be 18/12, which could be simplified to 6/4.
Hope this helps!
