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s344n2d4d5 [400]
3 years ago
11

Help asap please! 10 points, friend request, brainliest, thanks, 5 star rating for the first answer! 10 points, friend request,

thanks, 5 star rating for 2 answer!
I know D is incorrect.

Mathematics
2 answers:
Amanda [17]3 years ago
6 0

Answer:

Step-by-step explanation:

It's b (-2) and c (-1.2), Because A whole number can't be negative and an integer is a whole number and their opposites therefore it can't be something that can be written in a/b form,

hram777 [196]3 years ago
4 0

Answer:

-1.2

Step-by-step explanation:

An interger has to be a whole number that can be positive, negative, or zero. It can't have a decimal point. hope this helps

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Find the directional derivative of f(x,y,z)=z3−x2yf(x,y,z)=z3−x2y at the point (−5,5,2)(−5,5,2) in the direction of the vector v
olga_2 [115]

We are given

f=z^3 -x^2y

Firstly, we can find gradient

so, we will find partial derivatives

f_x=0 -2xy

f_x=-2xy

f_y=0 -x^2

f_y=-x^2

f_z=3z^2

now, we can plug point (-5,5,2)

f_x=-2*-5*5=50

f_y=-(-5)^2=-25

f_z=3(2)^2=12

so, gradient will be

gradf=(50,-25,12)

now, we are given that

it is in direction of v=⟨−3,2,−4⟩

so, we will find it's unit vector

|v|=\sqrt{(-3)^2+(2)^2+(-4)^2}

|v|=\sqrt{29}

now, we can find unit vector

v'=(\frac{-3}{\sqrt{29} } , \frac{2}{\sqrt{29} } , \frac{-4}{\sqrt{29} })

now, we can find dot product to find direction of the vector

dir=(gradf) \cdot (v')

now, we can plug values

dir=(50,-25,12) \cdot (\frac{-3}{\sqrt{29} } , \frac{2}{\sqrt{29} } , \frac{-4}{\sqrt{29} })

dir=(-\frac{150}{\sqrt{29} } - \frac{50}{\sqrt{29} } - \frac{48}{\sqrt{29} })

dir=-\frac{248\sqrt{29}}{29}.............Answer



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