All categories

2 years ago

I need help and fast I was supposed to turn this in last week

Mathematics

1 answer:

Reptile [31]2 years ago

3 0

Answer:

The answer is below

Step-by-step explanation:

If two lines intersected by a transversal form congruent corresponding angles, then the lines are parallel.

If two angles are supplementary to the same angle, then the two angles are congruent.

Statement Reason

1) line p is parallel to line q Given

2) angle 1 and angle 4 are Given

supplementary i.e. ∠1 + ∠4

= 180° (definition of supplementary)

3) angle 3 and angle 4 are consecutive interior angles

supplementary

4) angle 1 congruent to angle 3 congruent supplements theorem

supplementary

5) line r parallel to line s corresponding angle converse

You might be interested in

Please help me thank you

ladessa [460]

Answer:

Step-by-step explanation:

5 0

3 years ago

Which statement belongs in Step 4?

Gala2k [10]

Answer:

Step-by-step explanation:

TQ = QR

Ray t bisects TR. TR is divided into two equal parts

3 0

3 years ago

Consider the following region R and the vector field F. a. Compute the two-dimensional curl of the vector field. b. Evaluate bo

Shalnov [3]

Looks like we're given

$\vec F(x,y)=\langle-x,-y\rangle$

which in three dimensions could be expressed as

$\vec F(x,y)=\langle-x,-y,0\rangle$

and this has curl

$\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle$

which confirms the two-dimensional curl is 0.

It also looks like the region $R$ is the disk $x^2+y^2\le5$ . Green's theorem says the integral of $\vec F$ along the boundary of $R$ is equal to the integral of the two-dimensional curl of $\vec F$ over the interior of $R$ :

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA$

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of $R$ by

$\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle$

$\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt$

with $0\le t\le2\pi$ . Then

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt$