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andrew11 [14]
3 years ago
6

3/4 of what number is 51

Mathematics
1 answer:
NemiM [27]3 years ago
8 0

68


.75 of 68 is 51.

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Which number produces a rational number when multiplied by 0.5?
xxMikexx [17]

If the number begins by being rational, multiplying by 0.5 will not change whether or not it is rational. It will remain rational.


That would include any even number 12 * 0.5 = 6

12 started out as rational, so the answer is rational (6)


Any odd number is also rational

23 * 0.5 = 11.5

23 and 11.5 are both rational.


3.636363636 ... is rational so when multiplied by 0.5 the result will be rational.

1.818181818....


So if we start with something irrational like pi

then pi * 0.5 is not rational.

If you have choices, please list them.
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Step-by-step explanation:

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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)

For finding the right equation of the tangent plane, let's substitute the values of the partial derivatives evaluated at the given point:

z = -80 + 3x^2(x-1) - 4y^3(y-3)\\z = -80 +3(x-1)-108(y-3)\\z = 241 + 3 x - 108 y

or 3x-108y-z+241 = 0

7 0
3 years ago
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