Hi there!
We can calculate dy/dx using implicit differentiation:
xy + y² = 6
Differentiate both sides. Remember to use the Product Rule for the "xy" term:
(1)y + x(dy/dx) + 2y(dy/dx) = 0
Move y to the opposite side:
x(dy/dx) + 2y(dy/dx) = -y
Factor out dy/dx:
dy/dx(x + 2y) = -y
Divide both sides by x + 2y:
dy/dx = -y/x + 2y
We need both x and y to find dy/dx, so plug in the given value of x into the original equation:
-1(y) + y² = 6
-y + y² = 6
y² - y - 6 = 0
(y - 3)(y + 2) = 0
Thus, y = -2 and 3.
We can calculate dy/dx at each point:
At y = -2: dy/dx = -(-2) / -1+ 2(-2) = -2/5.
At y = 3: dy/dx = -(3) / -1 + 2(3) = -3/5.
Answer:
(1,-2)
Step-by-step explanation:
x= 1
y= 2
Answer: 3
Step-by-step explanation:
Set them equal because they are the same so 15x+5=16x+2 so x+2=5 so x=3
Answer:
The parenthesis need to be kept intact while applying the DeMorgan's theorem on the original equation to find the compliment because otherwise it will introduce an error in the answer.
Step-by-step explanation:
According to DeMorgan's Theorem:
(W.X + Y.Z)'
(W.X)' . (Y.Z)'
(W'+X') . (Y' + Z')
Note that it is important to keep the parenthesis intact while applying the DeMorgan's theorem.
For the original function:
(W . X + Y . Z)'
= (1 . 1 + 1 . 0)
= (1 + 0) = 1
For the compliment:
(W' + X') . (Y' + Z')
=(1' + 1') . (1' + 0')
=(0 + 0) . (0 + 1)
=0 . 1 = 0
Both functions are not 1 for the same input if we solve while keeping the parenthesis intact because that allows us to solve the operation inside the parenthesis first and then move on to the operator outside it.
Without the parenthesis the compliment equation looks like this:
W' + X' . Y' + Z'
1' + 1' . 1' + 0'
0 + 0 . 0 + 1
Here, the 'AND' operation will be considered first before the 'OR', resulting in 1 as the final answer.
Therefore, it is important to keep the parenthesis intact while applying DeMorgan's Theorem on the original equation or else it would produce an erroneous result.
I think It's $995 I HOPE THIS IS A CORRECT ANSWER
