What is the measure of n?

Please help me with this question. I will also give brainliest!

Alekssandra [29.7K]

What question? I don’t see no question

3 0

3 years ago

In a sale there is 25%off on all prices an TV costs £90 in the sale. How much was it before the sale?

Elina [12.6K]

To work out how much something was before the discount, you have to divide the price (which is £90) by 75 (as this is the percentage that you paid). It's gives you £1.20. Therefore, 1% is £1.20. To find the original cost, which is 100%, you have to multiply it by 100. Multiplying £1.20 by 100 gives you £120. Therefore, the cost of the TV before the 25% reeducation was £120. Hope this helps :)

3 0

4 years ago

Read 2 more answers

Trapezoid JKLM is shown on the coordinate plane below:

romanna [79]

L' would be (1, -4) so A

8 0

3 years ago

Read 2 more answers

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 $y^4$ )