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alexandr1967 [171]
3 years ago
7

Describe fully the single transformation that maps triangle A onto triangle B.

Mathematics
1 answer:
Brrunno [24]3 years ago
7 0

9514 1404 393

Answer:

  reflection across the line y = x

Step-by-step explanation:

The orientation of the triangle has changed from CW to CCW (considering sides in order least-to-greatest). This means a reflection is involved.

In order for this to be a single transformation, we need to know the line or point of reflection.

Each point is reflected about the point that is the midpoint joining an image point and the corresponding preimage point. Here, the points of reflection of the vertices all lie on the line y = x. So we can say ...

  the transform is reflection across the line y = x

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What is 4x-18 rewritten 4 different ways?
nasty-shy [4]
-18+4x
2(2x-9)
4x=18
-4x=-18
x=18/4
-x=-18/4
7 0
3 years ago
Which of the following expressions represents the distance between-1/4 and 3/4
Shalnov [3]

Answer:

|-1/4 - 3/4|

Step-by-step explanation:

To get the distance, take the absolute value of the second point minus the first point

| 3/4 - - 1/4|

|3/4 +  1/4|

We can also  take the absolute value of the first point minus the second point

|-1/4 - 3/4|

4 0
3 years ago
Work out the volume of the cuboid in cm³<br><br>​
Aliun [14]

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it is 192000 is got it right

8 0
3 years ago
What is the value of a? -2(a-8)+6a=36
Aleksandr [31]
First we expand the brackets:

-2a + 8 + 6a = 36
4a + 8 = 36

We then need to get the a's onto one side and the numbers onto the other:

4a = 28

We can then divide to get an x by itself:

a = 28/4 = 7
6 0
3 years ago
Read 2 more answers
A particle is moving along the x-axis so that its position at t ≥ 0 is given by s(t)=(t)ln(5t). Find the acceleration of the par
lyudmila [28]

Answer:

a(\frac{1}{5e})=5e

Step-by-step explanation:

we are given equation for position function as

s(t)=tln(5t)

Since, we have to find acceleration

For finding acceleration , we will find second derivative

s'(t)=\frac{d}{dt}\left(t\ln \left(5t\right)\right)

=\frac{d}{dt}\left(t\right)\ln \left(5t\right)+\frac{d}{dt}\left(\ln \left(5t\right)\right)t

=1\cdot \ln \left(5t\right)+\frac{1}{t}t

s'(t)=\ln \left(5t\right)+1

now, we can find derivative again

s''(t)=\frac{d}{dt}\left(\ln \left(5t\right)+1\right)

=\frac{d}{dt}\left(\ln \left(5t\right)\right)+\frac{d}{dt}\left(1\right)

=\frac{1}{t}+0

a(t)=\frac{1}{t}

Firstly, we will set velocity =0

and then we can solve for t

v(t)=s'(t)=\ln \left(5t\right)+1=0

we get

t=\frac{1}{5e}

now, we can plug that into acceleration

and we get

a(\frac{1}{5e})=\frac{1}{\frac{1}{5e}}

a(\frac{1}{5e})=5e


5 0
3 years ago
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