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3 years ago

What set of transformations are applied to parallelogram ABCD to create A'B'C'D'?

Mathematics

2 answers:

Verdich [7]3 years ago

7 0

Solution:

The information about the two parallelograms, one called Pre image (A B CD) and other called image (A'B'C'D').

Parallelogram formed by ordered pairs A at negative 4, 1, B at negative 3, 2, C at negative 1, 2, D at negative 2, 1. Second parallelogram transformed formed by ordered pairs A prime at negative 4, negative 1, B prime at negative 3, negative 2, C prime at negative 1, negative 2, D prime at negative 2, negative 1.

→As, the distance between X axis , that is distance of Preimage from X axis is same as distance of image from X axis.So,it is case of reflection over X axis.

→The two parallelogram ABCD and A'B'C'D' are congruent.So,Dilation hasn't taken place.Nor translation.

Option (B) →Reflected over the y-axis and rotated by 180° might be true.

pentagon [3]3 years ago

6 0

Answer: <u><em>Option (B)</em></u> →Reflected over the y-axis and rotated by 180° might be true.

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In the adjoining figure, XY = XZ . YQ and ZP are the bisectors of <img src="https://tex.z-dn.net/?f=%20%5Cangle" id="TexFormula1

svetlana [45]

Answer:

See Below.

Step-by-step explanation:

Statements: Reasons:

$1)\, XY=XZ$ Given

$2) \text{ $ m\angle Y= m\angle Z$}$ Isosceles Triangle Theorem

$\displaystyle 3) \text{ $m\angle Y=m\angle XYQ + \angle QYZ$}$ Angle Addition

$\displaystyle 4)\text{ $YQ$ bisects $\angle XYZ$}$ Given

$5) \text{ $m\angle XYQ=m\angle QYZ$}$ Definition of Bisector

$\displaystyle 6)\text{ $m\angle Y=2m\angle QYZ$}$ Substitution

$7)\text{ $m\angle Z=m\angle XZP+m\angle PZY$}$ Angle Addition

$8)\text{ $ZP$ bisects $\angle XZY$}$ Given

$\displaystyle 9) \text{ $m\angle XZP=m\angle PZY$ }$ Definition of Bisector

$\displaystyle 10) \text{ $ m\angle Z = 2m\angle PZY $}$ Substitution

$11)\text{ } 2m\angle QYZ=2m\angle PZY$ Substitution

$12)\text{ }m\angle QYZ=m\angle PZY$ Division Property of Equality

$13)\text{ } YZ=YZ$ Reflexive Property

$14)\text{ } \Delta YZP\cong\Delta ZYQ$ Angle-Side-Angle Congruence*

$15)\text{ } YQ=ZP$ CPCTC

*For clarification:

∠Y = ∠Z

YZ = YZ (or ZY)

∠PZY = ∠QYZ

So, Angle-Side-Angle Congruence:

ΔYZP is congruent to ΔZYQ

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3 years ago

Read 2 more answers

* Use order of operations to solve the following.

Bingel [31]

The answer is 100 if that 5 and 2 is separate

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Evaluate the surface integral:S

rjkz [21]

Assuming $S$ does not include the plane $z=0$ , we can parameterize the region in spherical coordinates using

$\mathbf r(u,v)=\left\langle3\cos u\sin v,3\sin u\sin v,3\cos v\right\rangle$

where $0\le u\le2\pi$ and $0\le v\le\dfrac\pi/2$ . We then have

$x^2+y^2=9\cos^2u\sin^2v+9\sin^2u\sin^2v=9\sin^2v$
$(x^2+y^2)=9\sin^2v(3\cos v)=27\sin^2v\cos v$

Then the surface integral is equivalent to

$\displaystyle\iint_S(x^2+y^2)z\,\mathrm dS=27\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^2v\cos v\left\|\frac{\partial\mathbf r(u,v)}{\partial u}\times \frac{\partial\mathbf r(u,v)}{\partial u}\right\|\,\mathrm dv\,\mathrm du$

We have

$\dfrac{\partial\mathbf r(u,v)}{\partial u}=\langle-3\sin u\sin v,3\cos u\sin v,0\rangle$
$\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle3\cos u\cos v,3\sin u\cos v,-3\sin v\rangle$
$\implies\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle-9\cos u\sin^2v,-9\sin u\sin^2v,-9\cos v\sin v\rangle$
$\implies\left\|\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}\|=9\sin v$

So the surface integral is equivalent to

$\displaystyle243\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv\,\mathrm du$
$=\displaystyle486\pi\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv$
$=\displaystyle486\pi\int_{w=0}^{w=1}w^3\,\mathrm dw$

where $w=\sin v\implies\mathrm dw=\cos v\,\mathrm dv$ .

$=\dfrac{243}2\pi w^4\bigg|_{w=0}^{w=1}$
$=\dfrac{243}2\pi$

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a_sh-v [17]

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yarga [219]

Step-by-step explanation:

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