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Advocard [28]
2 years ago
11

Please give a valid complete answer with steps. Given the information below, what is m ∠CBE?

Mathematics
1 answer:
Arte-miy333 [17]2 years ago
3 0

Answer:

42 + 5 x + 7x + 6 = 180 \\ 48 + 12x = 180 \\ 12x = 132 \\ x = 11 \\ 7x + 6 = 77 + 6 = 83

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Necesito ayuda por favor es para mañana
Elena L [17]

Answer: a. 14/30= 7/15

b. 16/30=8/15

C. 7/15= 47%

D.8/15= 53%

Step-by-step explanation:

3 0
3 years ago
Need help with math probelem i will give you 5 stars and a good rating
ArbitrLikvidat [17]

Answer:

8 feet

Step-by-step explanation:

12 x 2 =24

40 - 24 = 16

16 /2 = 8

Hope it helps :)

7 0
2 years ago
Find the quotient in SIMPLEST form:
Bas_tet [7]

Answer:Find the quotient in SIMPLEST form: D) 16

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Simplify.12x^3-24x^2/6x^3+12x
nordsb [41]

2+(-24x^2-24x)÷(12x^3-24x)

3 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)

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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