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natka813 [3]
3 years ago
15

On a coordinate plane, 3 triangles are shown. Triangle B C D has points (1, 4), (1, 2), (5, 3). Triangle B prime C prime D prime

has points (negative 1, 4), (negative 1, 2), (negative 5, 3). Triangle B double-prime C double-prime D double-prime has points (5, negative 1), (5, negative 3), (1, negative 2). Which rule describes the composition of transformations that maps ΔBCD to ΔB"C"D"? Translation of 5 units x, negative 6 units y composition reflection across y = negative x Reflection across y = negative x composition translation of 5 units x, negative 6 units y. Translation of 6 units x, negative 5 units y composition reflection across the y-axis Reflection across the y-axis composition translation of 6 units x, negative 5 units y
Mathematics
2 answers:
wel3 years ago
8 0

Answer:

Its c

Explanation:

I got it right on the test

Tresset [83]3 years ago
3 0

The rule which describe the composition of transformations that

maps ΔBCD to ΔB"C"D" is:

Reflection across the y-axis composition translation of 6 units x,

negative 5 units y ⇒ last answer

Step-by-step explanation:

Let us revise the reflection across the y-axis , horizontal translation

and vertical translation

1. If point (x , y) is reflected across the y-axis, then its image is (-x , y)

2. If point (x , y) is translated h units to the right, then its image is

   (x + h , y), if translated h units to the left, then its image is (x - h , y)

3. If point (x , y) is translated k units up, then its image is (x , y + k),

   if translated k units down, then its image is (x , y - k)

∵ The vertices of triangle BCD are (1 , 4) , (1 , 2) , (5 , 3)

∵ The vertices of triangle B'C'D' are (-1 , 4) , (-1 , 2) , (-5 , 3)

∵ The x-coordinates of the vertices of Δ B'C'D' have the same

   magnitude of x-coordinates of Δ BCD and opposite signs

∴ Δ B'C'D' is the image of Δ ABC after reflection across the y-axis

∵ The vertices of triangle B'C'D' are (-1 , 4) , (-1 , 2) , (-5 , 3)

∵ The vertices of triangle B''C''D'' are (5 , -1) , (5 , -3) , (1 , -2)

∵ The image of -1 is 5 and the image of -5 is 1

∴ The x-coordinates of the vertices of triangle B'C'D' are added by 6

∵ The image of 4 is -1 , image of 2 is -3 and the image of 3 is -2

∴ The y-coordinates of the vertices of triangle B'C'D' are subtracted

   by 5

∴ Δ B"C"D" is the image of Δ B'C'D' by translate 6 units to the right

  and 5 units down ⇒ (x + 6 , y - 5)

The rule which describe the composition of transformations that

maps ΔBCD to ΔB"C"D" is:

Reflection across the y-axis composition translation of 6 units x,

negative 5 units y

Learn more:

You can learn more about reflection in brainly.com/question/11203617

#LearnwithBrainly

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Performance Matters
spayn [35]

Answer:

the 1st, 2nd and 6th statements are true.

and probably the 3rd statement is true.

I am not sure about the 3rd statement, as I cannot read the original in your screenshot, and your transcribed description is probably not correct and contains typos.

but all you need is in the explanation below to decide, if the actual 3rd statement is true or not.

if "58 + 79 = 1268" actually means "5s + 7g = 1268" then it is true. otherwise it is false.

Step-by-step explanation:

g = number of general tickets sold

s = number of student tickets sold

so, in total

g + s = 234

tickets were sold.

and the revenue was

7g + 5s = $1,268

out of these 2 basic equating we get

g = 234 - s

and then

7(234 - s) + 5s = 1,268

1,638 - 7s + 5s = 1,268

-2s = -370

s = 185

g = 234 - s = 234 - 185 = 49

so, we know

185 student tickets were sold for 5×185 = $925

49 general tickets were sold for 7×49 = $343

7 0
2 years ago
Having trouble <br> With this math question just gotta this unit
Scorpion4ik [409]
What are you suppose to do - solve it or reduce it?  
3 0
3 years ago
Read 2 more answers
Find the volume of the figure.
liberstina [14]
A triangular prisms' formula is: lwh = Length * Width * Height
substitute the variables
L = 21in
W = 15in
H = 17in
21*15*17
= 5,355in

The volume of the figure is: 5,355in^3


7 0
3 years ago
Help please thanks :)
Gre4nikov [31]

Answer:

the third one down, there is no gap in the data set

Step-by-step explanation:

ill be here all day

8 0
3 years ago
Use the divergence theorem to calculate the surface integral s f · ds; that is, calculate the flux of f across s. f(x, y, z) = x
valkas [14]
\mathbf f(x,y,z)=x^4\,\mathbf i-x^3z^2\,\mathbf j+4xy^2z\,\mathbf k
\mathrm{div}(\mathbf f)=\dfrac{\partial(x^4)}{\partial x}+\dfrac{\partial(-x^3z^2)}{\partial y}+\dfrac{\partial(4xy^2z)}{\partial z}=4x^3+0+4xy^2=4x(x^2+y^2)


Let \mathcal D be the region whose boundary is \mathcal S. Then by the divergence theorem,

\displaystyle\iint_{\mathcal S}\mathbf f\cdot\mathrm d\mathbf S=\iiint_{\mathcal D}4x(x^2+y^2)\,\mathrm dV

Convert to cylindrical coordinates, setting

x=r\cos\theta
y=r\sin\theta

and keeping z as is. Then the volume element becomes


\mathrm dV=r\,\mathrm dr\,\mathrm d\theta\,\mathrm dz

and the integral is

\displaystyle\iiint_{\mathcal D}4x(x^2+y^2)\,\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=0}^{z=r\cos\theta+7}4r\cos\theta\cdot r^2\cdot r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta
=\displaystyle4\iiint_{\mathcal D}r^4\cos\theta\,\mathrm dz\,\mathrm dr\,\mathrm d\theta
=\dfrac{2\pi}3
4 0
4 years ago
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