The table shows the lowest temperatures recorded on five days in April 2020, but the value for 4/23/20 has been lost . if X Is t
he lost value , and we know that X was the middle value of the five days , which choice could be the value of X?
A. -1.9 °C
B. -0.9 °C
C. -0.1 °C
D. -0.4 °C
A. -1.9 °C
B. -0.9 °C
C. -0.1 °C
D. -0.4 °C
2 answers:
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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.