Correct and best explained answer gets brainliest.

Sergeu [11.5K]

Answer:

Step-by-step explanation:

125 is a composite number. 125 = 1 x 125 or 5 x 25. Factors of 125: 1, 5, 25, 125. Prime factorization: 125 = 5 x 5 x 5, which can also be written 125 = 5³.

3 0

3 years ago

Select the domain and range of F.

Strike441 [17]

Answer:

b

Step-by-step explanation:

4 0

3 years ago

A student was asked to find the equation of the tangent plane to the surface z=x3−y4 at the point (x,y)=(1,3). The student's ans

aliya0001 [1]

Answer:

C) The partial derivatives were not evaluated a the point.

D) The answer is not a linear function.

The correct equation for the tangent plane is $z = 241 + 3 x - 108 y$ or $3x-108y-z+241 = 0$

Step-by-step explanation:

The equation of the tangent plane to a surface given by the function $S=f(x, y)$ in a given point $(x_0, y_0, z_0)$ can be obtained using:

$z-z_0=f_x(x_0,y_0,z_0)(x-x_0)+f_y(x_0,y_0, z_0)(y-y_0)$ (1)

where $f_x(x_0, y_0, z_0)$ and $f_y(x_0, y_0, z_0)$ are the partial derivatives of $f(x,y)$ with respect to $x$ and $y$ respectively and evaluated at the point $(x_0, y_0, z_0)$ .

Therefore we need to find two missing inputs in our problem in order to use equation (1). The $z_0$ coordinate and the partial derivatives $f_x(x_0, y_0, z_0)$ and $f_y(x_0, y_0, z_0)$ . For $z_0$ just evaluating in the given function we obtain $z_0= -80$ and the partial derivatives are:

$\frac{\partial f(x,y)}{\partial x} \equiv f_x(x, y)= 3x^2 \\f_x(x_0, y_0) = f_x(1, 3) = 3$

$\frac{\partial f(x,y)}{\partial y} \equiv f_y(x, y)= -4y^3 \\f_y(x_0, y_0) = f_y(1, 3) = -108$

Now, substituting in (1)

$z-z_0=f_x(x_0,y_0,z_0)(x-x_0)+f_y(x_0,y_0, z_0)(y-y_0)\\z + 80 = 3x^2(x-1) - 4y^3(y-3)\\z = -80 + 3x^2(x-1) - 4y^3(y-3)$

Notice that until this point, we obtain the same equation as the student, however, we have not evaluated the partial derivatives and therefore this is not the equation of the plane and this is not a linear function because it contains the terms ( $x^3$ and $y^4$ )