The initial value of a function is the y value when x = 0. For the first function, you can plug in the value of x =0 and you'll get 2/3. Function 2 has an initial value of 3 (you can look at the table). Since 3 > 2/3, function 2 has a greater initial value.
Note: I'm guessing that the table refers to Function 2.
Answer:
3.
Step-by-step explanation:
Implicit differentiation:
x^2 y + (xy)^3 + 3x = 0
x^2 y + x^3y^3 + 3x = 0
Using the product rule:
2x* y + x^2*dy/dx + 3x^2 y^3 + x^3* (d(y^3)/dx) + 3 = 0
2xy + x^2 dy/dx + 3x^2 y^3 + x^3* 3y^2 dy/dx + 3 = 0
dy/dx(x^2 + 3y^2x^3) = (-2xy - 3x^2y^3 - 3)
dy/dx= (-2xy - 3x^2y^3 - 3) / (x^2 + 3y^2x^3)
At the point (-1, 3).
the derivative = (6 - 81 - 3)/(1 -27)
= -78/-26
= 3.
The only other possible meaning I can see is that a2 + b2 means a^2 + b^2. In the case that is a formula that cannot be simplified further, although it could be written as (a + b)^2 - 2ab or (a - b)^2 + 2ab. ... Originally Answered: What is the formula of a2+b2?
C) Both (house, complete mailing address) and (complete mailing address, house) are functions.
I think the answer is 716.7 not 100% positive