I dont speak your lingo m8
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Answer:
Relation 1: Not a Function
Relation 2: Not a Function
Relation 3: Function
Relation 4: Not a Function
Explanation:
There can only be one output (what you get out of it) for each input (what you put into the function) in a function.
Answer:
the answer is C. because the factor was 36x^3-15x^2-6x so i got 3(x^2+1)(4×-1)
Answer:
three consecutive odd integers: 2n-1 2n+1 2n+3
that the sum of the smaller two is three
times larger increased by seven: 2n-1 + 2n+1 = 3(2n+3) +7
4n = 6n+ 9 +7
4n = 6n+ 16
4n -6n = 16
-2n = 16
n = 16/(-2)
n=-8
a) 2n -1 = 2(-8) -1 = -17
b) 2n+1 = 2(-8)+1 = -15
c) 2n+1 = 2(-8)+3 = -13
Ans. -17 ; -15 ; -13
I'm reading this as

with

.
The value of the integral will be independent of the path if we can find a function

that satisfies the gradient equation above.
You have

Integrate

with respect to

. You get


Differentiate with respect to

. You get
![\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}[x^2e^{-y}+g(y)]](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20y%7D%3D%5Cdfrac%7B%5Cpartial%7D%7B%5Cpartial%20y%7D%5Bx%5E2e%5E%7B-y%7D%2Bg%28y%29%5D)


Integrate both sides with respect to

to arrive at



So you have

The gradient is continuous for all

, so the fundamental theorem of calculus applies, and so the value of the integral, regardless of the path taken, is