All categories

3 years ago

Which segment is congruent to segment GJ

1 answer:

7 0

<span>well congruent means they must be equal so since from point G to point J is -3 spaces you would just need to find another segment that is also -3 spaces, JM. For the second question i think you mean find the midpoint between NH, and if so you would need to figure out 0 - 6, which gives you -3 which is located at point K. Hope this is the right one.</span>

You might be interested in

What is 449/60 as a mixed number

Ilia_Sergeevich [38]

$\dfrac{449}{60}=\dfrac{300+120+29}{60}=\dfrac{300}{60}+\dfrac{120}{60}+\dfrac{29}{60}\\\\=5+2+\dfrac{29}{60}=\boxed{7\dfrac{29}{60}}$

6 0

4 years ago

<img src="https://tex.z-dn.net/?f=%5Ctext%7BSolve%20for%20x%3A%7D%5C%5C%5C%5C3x%20%2B%2054%20%3D%20180" id="TexFormula1" title="

vagabundo [1.1K]

Hi there!

»»————-　★　————-««

I believe your answer is:

$x = 42$

»»————-　★　————-««

Here’s why:

We will use inverse operations to solve for 'x'.

⸻⸻⸻⸻

$\boxed{\text{Calculating the Answer....}}\\\\3x+54=180\\----------\\\rightarrow 3x + 54 - 54 = 180-54\\\\\rightarrow 3x = 126\\\\\rightarrow\frac{3x=126}{3}\\\\\rightarrow\boxed{x=42}$

⸻⸻⸻⸻

»»————-　★　————-««

Hope this helps you. I apologize if it’s incorrect.

3 0

3 years ago

Read 2 more answers

Please Help Me with this problem

Alex Ar [27]

Answer:

AEB =305° as stated below

Step-by-step explanation:

AEB is the larger area starting A to EB and missing out the small AB area

EC = 180° = supplementary we label 87° angle to P.

P+ BCD = 87, + 38 = 125°

ABCD = 180°

180 - BCD = 180-125°

AB = 55°

Therefore AEB = 360-55=305°

3 0

3 years ago

Calcule el siguiente sueldo: Sueldo base: $934.55 Horas al 50%: 9 Fondo de reserva: si Decimos proporcionales: si Afiliación al

Masja [62]

Answer:

I speak some spanish but this i can not understand

Step-by-step explanation:

6 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