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4 years ago

Lines l and m are parallel and intersected by transversals t and s as shown in the figure below.

Mathematics

1 answer:

kogti [31]4 years ago

8 0

Angle W and Y are right angles and equal to 90 degrees each. Angle z and angle x are same side interior angles and they have a sum of 180 degrees because of the same side interior angles postulate.

so W= 90 Y=90 X+Z= 180 . so 90+180+90 =360 degrees. B.

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From the graph, write the linear equation in slope-intercept form

elena55 [62]

Answer:

1.y=1/2+2

2.y=-2+1

3.y=3/2+3

4.y=-2/3-2

Step-by-step explanation:

4 0

3 years ago

To solve the system of linear equations 3 x minus 2 y = 4 and 9 x minus 6 y = 12 by using the linear combination method, Henry d

seraphim [82]

Answer:

Step-by-step explanation:

Both lines have same slope so the graph of both equation will be same and hence it will overlap each other that is why the Henry could only see one line.

Step-by-step explanation:

Given : A system of linear equation 3x - 2y = 4 and 9x - 6y = 12

We have to show whey when Henry graphed the equations 3x-2y=4 and 9x-6y=12 he could only see one line.

Consider the given system of linear equation

3x - 2y = 4 ................(1)

9x - 6y = 12 ..................(2)

Since,

Equation (2) is multiple of equation (1),

3 × ( 3x - 2y = 4) = 9x - 6y = 12

Also the slope of given equations are same

For equation (1),

Differentiate with respect to x, we get,

3-2dy/dx = 0

dy/dx = 3/2

Also for equation (2),

Differentiate with respect to x, we get,

9 - 6dy/dx = 0

dy/dx = 9/6 = 3/2

Since, both lines have same slope so the graph of both equation will be same and hence it will overlap each other.

That is why the Henry could only see one line.

3 0

2 years ago

find an equation of the tangent plane to the given parametric surface at the specified point. x = u v, y = 2u2, z = u − v; (2, 2

Alexxandr [17]

The surface is parameterized by

$\vec s(u,v) = x(u, v) \, \vec\imath + y(u, v) \, \vec\jmath + z(u, v)) \, \vec k$

and the normal to the surface is given by the cross product of the partial derivatives of $\vec s$ :

$\vec n = \dfrac{\partial \vec s}{\partial u} \times \dfrac{\partial \vec s}{\partial v}$

It looks like you're given

$\begin{cases}x(u, v) = u + v\\y(u, v) = 2u^2\\z(u, v) = u - v\end{cases}$

Then the normal vector is

$\vec n = \left(\vec\imath + 4u \, \vec\jmath + \vec k\right) \times \left(\vec \imath - \vec k\right) = -4u\,\vec\imath + 2 \,\vec\jmath - 4u\,\vec k$